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Section 1.5 The Tangent Line Approximation

Among all functions, linear functions are simplest. One of the powerful consequences of a function \(y = f(x)\) being differentiable at a point \((a,f(a))\) is that, up close, the function \(y = f(x)\) is locally linear and looks like its tangent line at that point. In certain circumstances, this allows us to approximate the original function \(f\) with a simpler function \(L\) that is linear: this can be advantageous when we have limited information about \(f\) or when \(f\) is computationally or algebraically complicated. We will explore all of these situations in what follows.

It is essential to recall that when \(f\) is differentiable at \(x = a\text{,}\) the value of \(f'(a)\) provides the slope of the tangent line to \(y = f(x)\) at the point \((a,f(a))\text{.}\) If we know both a point on the line and the slope of the line we can find the equation of the tangent line and write the equation in point-slope form 1 .

Recall that a line with slope \(m\) that passes through \((x_0,y_0)\) has equation \(y - y_0 = m(x - x_0)\text{,}\) and this is the point-slope form of the equation.
Preview Activity 1.5.1.

Consider the function \(y = g(x) = -x^2+3x+2\text{.}\)

  1. Use the limit definition of the derivative to compute a formula for \(y = g'(x)\text{.}\)

  2. Determine the slope of the tangent line to \(y = g(x)\) at the value \(x = 2\text{.}\)

  3. Compute \(g(2)\text{.}\)

  4. Find an equation for the tangent line to \(y = g(x)\) at the point \((2,g(2))\text{.}\) Write your result in point-slope form.

  5. On the axes provided in Figure 1.5.1, sketch an accurate, labeled graph of \(y = g(x)\) along with its tangent line at the point \((2,g(2))\text{.}\)

Figure 1.5.1. Axes for plotting \(y = g(x)\) and its tangent line to the point \((2,g(2))\text{.}\)

Subsection 1.5.1 The tangent line

Given a function \(f\) that is differentiable at \(x = a\text{,}\) we know that we can determine the slope of the tangent line to \(y = f(x)\) at \((a,f(a))\) by computing \(f'(a)\text{.}\) The equation of the resulting tangent line is given in point-slope form by

\begin{equation*} y - f(a) = f'(a)(x-a) \ \ \text{or} \ \ y = f'(a)(x-a) + f(a)\text{.} \end{equation*}

Note well: there is a major difference between \(f(a)\) and \(f(x)\) in this context. The former is a constant that results from using the given fixed value of \(a\text{,}\) while the latter is the general expression for the rule that defines the function. The same is true for \(f'(a)\) and \(f'(x)\text{:}\) we must carefully distinguish between these expressions. Each time we find the tangent line, we need to evaluate the function and its derivative at a fixed \(a\)-value.

In Figure 1.5.2, we see the graph of a function \(f\) and its tangent line at the point \((a,f(a))\text{.}\) Notice how when we zoom in we see the local linearity of \(f\) more clearly highlighted. The function and its tangent line are nearly indistinguishable up close. Local linearity can also be seen dynamically in the java applet at http://gvsu.edu/s/6J.

Figure 1.5.2. A function \(y = f(x)\) and its tangent line at the point \((a,f(a))\text{:}\) at left, from a distance, and at right, up close. At right, we label the tangent line function by \(y = L(x)\) and observe that for \(x\) near \(a\text{,}\) \(f(x) \approx L(x)\text{.}\)

Subsection 1.5.2 The local linearization

A slight change in perspective and notation will enable us to be more precise in discussing how the tangent line approximates \(f\) near \(x = a\text{.}\) By solving for \(y\text{,}\) we can write the equation for the tangent line as

\begin{equation*} y = f'(a)(x-a) + f(a) \end{equation*}

This line is itself a function of \(x\text{.}\) Replacing the variable \(y\) with the expression \(L(x)\text{,}\) we call

\begin{equation*} L(x) = f'(a)(x-a) + f(a) \end{equation*}

the local linearization of \(f\) at the point \((a,f(a))\text{.}\) In this notation, \(L(x)\) is nothing more than a new name for the tangent line. As we saw above, for \(x\) close to \(a\text{,}\) \(f(x) \approx L(x)\text{.}\)

Suppose that a function \(y = f(x)\) has its tangent line approximation given by \(L(x) = 3 - 2(x-1)\) at the point \((1,3)\text{,}\) but we do not know anything else about the function \(f\text{.}\) To estimate a value of \(f(x)\) for \(x\) near 1, such as \(f(1.2)\text{,}\) we can use the fact that \(f(1.2) \approx L(1.2)\) and hence

\begin{equation*} f(1.2) \approx L(1.2) = 3 - 2(1.2-1) = 3 - 2(0.2) = 2.6\text{.} \end{equation*}

We emphasize that \(y = L(x)\) is simply a new name for the tangent line function. Using this new notation and our observation that \(L(x) \approx f(x)\) for \(x\) near \(a\text{,}\) it follows that we can write

\begin{equation*} f(x) \approx f(a) + f'(a)(x-a) \ \text{for} \ x \ \text{near} \ a\text{.} \end{equation*}
Activity 1.5.2.

Suppose it is known that for a given differentiable function \(y = g(x)\text{,}\) its local linearization at the point where \(a = -1\) is given by \(L(x) = -2 + 3(x+1)\text{.}\)

  1. Compute the values of \(L(-1)\) and \(L'(-1)\text{.}\)

  2. What must be the values of \(g(-1)\) and \(g'(-1)\text{?}\) Why?

  3. Do you expect the value of \(g(-1.03)\) to be greater than or less than the value of \(g(-1)\text{?}\) Why?

  4. Use the local linearization to estimate the value of \(g(-1.03)\text{.}\)

  5. Suppose that you also know that \(g''(-1) = 2\text{.}\) What does this tell you about the graph of \(y = g(x)\) at \(a = -1\text{?}\)

  6. For \(x\) near \(-1\text{,}\) sketch the graph of the local linearization \(y = L(x)\) as well as a possible graph of \(y = g(x)\) on the axes provided in Figure 1.5.4.

Figure 1.5.4. Axes for plotting \(y = L(x)\) and \(y = g(x)\text{.}\)
Hint

  1. Follow the rule for \(L\text{.}\)

  2. Recall that the form of the local linearization is \(L(x) = g(a) + g'(a)(x-a)\text{.}\)

  3. Is the function \(g\) increasing or decreasing at \(a = -1\text{?}\)

  4. Remember that \(g(-1.03) \approx L(-1.03)\text{.}\)

  5. What does the second derivative tell you about the shape of a curve?

  6. Use your work above.

Answer

  1. \(L(-1) = -2\text{;}\) \(L'(-1) = 3\text{.}\)

  2. \(g(-1) = -2\text{;}\) \(g'(-1) = 3\text{.}\)

  3. Increasing.

  4. O\(g(-1.03) \approx L(-1.03) = -2.09\text{.}\)

  5. Concave up.

  6. The illustration below shows a possible graph of \(y = g(x)\) near \(x = -1\text{,}\) along with the tangent line \(y = L(x)\) through \((-1, g(-1))\text{.}\)

Solution

  1. Using the formula for \(L\text{,}\) \(L(-1) = -2\text{;}\) Since \(L'(x) = 3\text{,}\) we see \(L'(-1) = 3\text{.}\)

  2. Since \(L(x) = g(-1) + g'(-1)(x+1) = -2 + 3(x+1)\text{,}\) we see \(g(-1) = -2\) and \(g'(-1) = 3\text{.}\) Alternatively, we could observe that the value and slope of \(g\) must match the value and slope of \(L\) at the point of tangency.

  3. Because \(g'(-1) = 3\text{,}\) we see that \(g\) is increasing at \(a = -1\text{?}\)

  4. Observe that \(g(-1.03) \approx L(-1.03) = -2 + 3(-1.03+1) = -2 - 0.09 = -2.09\text{.}\)

  5. Since \(g''(-1) > 0\text{,}\) we know \(g\) is concave up at \(x = -1\text{.}\)

  6. In the figure below, we use the results of our previous work to generate the plot shown, which is a possible graph of \(y = g(x)\) near \(x = -1\text{,}\) along with the tangent line \(y = L(x)\) through \((-1, g(-1))\text{.}\)

From Activity 1.5.2, we see that the local linearization \(y = L(x)\) is a linear function that shares two important values with the function \(y = f(x)\) that it is derived from. In particular,

  • because \(L(x) = f(a) + f'(a)(x-a)\text{,}\) it follows that \(L(a) = f(a)\text{;}\) and
  • because \(L\) is a linear function, its derivative is its slope.

Hence, \(L'(x) = f'(a)\) for every value of \(x\text{,}\) and specifically \(L'(a) = f'(a)\text{.}\) Therefore, we see that \(L\) is a linear function that has both the same value and the same slope as the function \(f\) at the point \((a,f(a))\text{.}\)

Thus, if we know the linear approximation \(y = L(x)\) for a function, we know the original function's value and its slope at the point of tangency. What remains unknown, however, is the shape of the function \(f\) at the point of tangency. There are essentially four possibilities, as shown in Figure 1.5.5.

Figure 1.5.5. Four possible graphs for a nonlinear differentiable function and how it can be situated relative to its tangent line at a point.

These possible shapes result from the fact that there are three options for the value of the second derivative: either \(f''(a) \lt 0\text{,}\) \(f''(a) = 0\text{,}\) or \(f''(a) \gt 0\text{.}\)

  • If \(f''(a) \gt 0\text{,}\) then we know the graph of \(f\) is concave up, and we see the first possibility on the left, where the tangent line lies entirely below the curve.
  • If \(f''(a) \lt 0\text{,}\) then \(f\) is concave down and the tangent line lies above the curve, as shown in the second figure.
  • If \(f''(a) = 0\) and \(f''\) changes sign at \(x = a\text{,}\) the concavity of the graph will change, and we will see either the third or fourth figure. 2 .
  • A fifth option (which is not very interesting) can occur if the function \(f\) itself is linear, so that \(f(x) = L(x)\) for all values of \(x\text{.}\)
It is possible that \(f''(a) = 0\) but \(f''\) does not change sign at \(x = a\text{,}\) in which case the graph will look like one of the first two options.

The plots in Figure 1.5.5 highlight yet another important thing that we can learn from the concavity of the graph near the point of tangency: whether the tangent line lies above or below the curve itself. This is key because it tells us whether or not the tangent line approximation's values will be too large or too small in comparison to the true value of \(f\text{.}\) For instance, in the first situation in the leftmost plot in Figure 1.5.5 where \(f''(a) > 0\text{,}\) because the tangent line falls below the curve, we know that \(L(x) \le f(x)\) for all values of \(x\) near \(a\text{.}\)

Activity 1.5.3.

This activity concerns a function \(f(x)\) about which the following information is known:

  • \(f\) is a differentiable function defined at every real number \(x\)

  • \(f(2) = -1\)

  • \(y = f'(x)\) has its graph given in Figure 1.5.6

Figure 1.5.6. At center, a graph of \(y = f'(x)\text{;}\) at left, axes for plotting \(y = f(x)\text{;}\) at right, axes for plotting \(y = f''(x)\text{.}\)

Your task is to determine as much information as possible about \(f\) (especially near the value \(a = 2\)) by responding to the questions below.

  1. Find a formula for the tangent line approximation, \(L(x)\text{,}\) to \(f\) at the point \((2,-1)\text{.}\)

  2. Use the tangent line approximation to estimate the value of \(f(2.07)\text{.}\) Show your work carefully and clearly.

  3. Sketch a graph of \(y = f''(x)\) on the righthand grid in Figure 1.5.6; label it appropriately.

  4. Is the slope of the tangent line to \(y = f(x)\) increasing, decreasing, or neither when \(x = 2\text{?}\) Explain.

  5. Sketch a possible graph of \(y = f(x)\) near \(x = 2\) on the lefthand grid in Figure 1.5.6. Include a sketch of \(y=L(x)\) (found in part (a)). Explain how you know the graph of \(y = f(x)\) looks like you have drawn it.

  6. Does your estimate in (b) over- or under-estimate the true value of \(f(2.07)\text{?}\) Why?

Hint

  1. Find the value of \(f'(2)\) from the given graph of \(f\text{.}\)

  2. Remember that \(f(2.07) \approx L(2.07)\text{.}\)

  3. The graph of \(y = f''(x)\) is the derivative of the graph of \(y = f'(x)\text{.}\)

  4. Is \(f'\) increasing, decreasing, or neither when \(x = 2\text{?}\)

  5. Draw \(y = L(x)\) first. Then think about options for \(f\) relative to the graph of \(L\text{.}\)

  6. Does the tangent line lie above or below the graph of \(y = f(x)\) at \((2,3)\text{?}\)

Answer

  1. \(L(x) = -1 + 2(x-2)\text{.}\)

  2. \(f(2.07) \approx L(2.07) = -0.86\text{.}\)

  3. See the image in part e.

  4. Neither.

  5. See the image below, which shows, at left, a possible graph of \(y = f(x)\) near \(x = 2\text{,}\) along with the tangent line \(y = L(x)\) through \((2, f(2))\text{.}\)

  6. Too large.

Solution

  1. Since \(f(2) = -1\) and \(f'(2) = 2\text{,}\) we have \(L(x) = -1 + 2(x-2)\text{.}\)

  2. Using our work in (a), \(f(2.07) \approx L(2.07) = -1 + 2(2.07-2) = -1 + 2\cdot 0.07 = -0.86\text{.}\)

  3. See the image in part e.

  4. The slope of the tangent line to \(y = f(x)\) is increasing for \(x \lt 2\) because \(y = f'(x)\) is an increasing function on this interval. Similarly, for \(x > 2\text{,}\) the slope of the tangent line to \(y = f(x)\) is decreasing. Right at \(x = 2\text{,}\) the slope of the tangent line to \(y = f(x)\) is neither increasing nor decreasing.

  5. See the plot in the image below, which shows, at left, a possible graph of \(y = f(x)\) near \(x = 2\text{,}\) along with the tangent line \(y = L(x)\) through \((2, f(2))\text{.}\) Note that \(y = f(x)\) is concave up for \(x \lt 2\) since \(f'\) is increasing on that interval, and \(y = f(x)\) is concave down for \(x > 2\) since \(f'\) is decreasing there. Hence \(y = f(x)\) changes from concave up to concave down right at \(x = 2\text{,}\) which is also the point near 2 where the graph of \(y = f(x)\) is steepest.

  6. Because the tangent line to \(y = f(x)\) lies above the graph of \(f\) to the right of \(x = 2\text{,}\) our estimate of \(f(2.07)\) is too large — the local linearization overshoots the true value of \(f\) at this point.

The idea that a differentiable function looks linear and can be well-approximated by a linear function is an important one that finds wide application in calculus. For example, by approximating a function with its local linearization, it is possible to develop an effective algorithm to estimate the zeroes of a function. Local linearity also helps us to make further sense of certain challenging limits. For instance, we have seen that the limit

\begin{equation*} \lim_{x \to 0} \frac{\sin(x)}{x} \end{equation*}

is indeterminate, because both its numerator and denominator tend to 0. While there is no algebra that we can do to simplify \(\frac{\sin(x)}{x}\text{,}\) it is straightforward to show that the linearization of \(f(x) = \sin(x)\) at the point \((0,0)\) is given by \(L(x) = x\text{.}\) Hence, for values of \(x\) near 0, \(\sin(x) \approx x\text{,}\) and therefore

\begin{equation*} \frac{\sin(x)}{x} \approx \frac{x}{x} = 1\text{,} \end{equation*}

which makes plausible the fact that

\begin{equation*} \lim_{x \to 0} \frac{\sin(x)}{x} = 1\text{.} \end{equation*}

Subsection 1.5.3 Summary

  • The tangent line to a differentiable function \(y = f(x)\) at the point \((a,f(a))\) is given in point-slope form by the equation

    \begin{equation*} y - f(a) = f'(a)(x-a)\text{.} \end{equation*}

  • The principle of local linearity tells us that if we zoom in on a point where a function \(y = f(x)\) is differentiable, the function will be indistinguishable from its tangent line. That is, a differentiable function looks linear when viewed up close. We rename the tangent line to be the function \(y = L(x)\text{,}\) where \(L(x) = f(a) + f'(a)(x-a)\text{.}\) Thus, \(f(x) \approx L(x)\) for all \(x\) near \(x = a\text{.}\)

  • If we know the tangent line approximation \(L(x) = f(a) + f'(a)(x-a)\) to a function \(y=f(x)\text{,}\) then because \(L(a) = f(a)\) and \(L'(a) = f'(a)\text{,}\) we also know the values of both the function and its derivative at the point where \(x = a\text{.}\) In other words, the linear approximation tells us the height and slope of the original function. If, in addition, we know the value of \(f''(a)\text{,}\) we then know whether the tangent line lies above or below the graph of \(y = f(x)\text{,}\) depending on the concavity of \(f\text{.}\)

Exercises 1.5.4 Exercises

1. Approximating \(\sqrt{x}\).
2. Local linearization of a graph.
3. Estimating with the local linearization.
4. Predicting behavior from the local linearization.
5.

A certain function \(y=p(x)\) has its local linearization at \(a = 3\) given by \(L(x) = -2x + 5\text{.}\)

  1. What are the values of \(p(3)\) and \(p'(3)\text{?}\) Why?

  2. Estimate the value of \(p(2.79)\text{.}\)

  3. Suppose that \(p''(3) = 0\) and you know that \(p''(x) \lt 0\) for \(x \lt 3\text{.}\) Is your estimate in (b) too large or too small?

  4. Suppose that \(p''(x) \gt 0\) for \(x \gt 3\text{.}\) Use this fact and the additional information above to sketch an accurate graph of \(y = p(x)\) near \(x = 3\text{.}\) Include a sketch of \(y = L(x)\) in your work.

Answer

  1. \(p(3) = -1\) and \(p'(3) = -2\text{.}\)

  2. \(p(2.79) \approx -0.58\text{.}\)

  3. Too large.

Solution

  1. Since the value and slope of the tangent line match the value and slope of the function at the point of tangency, we know \(p(3) = L(3) = -1\) and \(p'(3) = L'(3) = -2\text{.}\)

  2. Using the local linearization, \(p(2.79) \approx L(2.79) = -2(2.79) + 5 = -0.58\text{.}\)

  3. Since \(p''(x) \lt 0\) for \(x \lt 3\text{,}\) we know that \(p\) is concave down for \(x \lt 3\text{.}\) This means to the left of the point of tangency at \(x = 3\text{,}\) the tangent line lies above the curve, and thus the estimate in (b) is too large.

  4. In the following figure, we see a possible function \(p\) that matches the value and slope of the tangent line at the point \((3,-1)\) and that is concave down to the left of \(x=3\) and concave up to the right.

6.

A potato is placed in an oven, and the potato's temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes.

\(t\) \(F(t)\)
\(0\) \(70\)
\(15\) \(180.5\)
\(30\) \(251\)
\(45\) \(296\)
\(60\) \(324.5\)
\(75\) \(342.8\)
\(90\) \(354.5\)
Table 1.5.7. Temperature data for the potato.

  1. Use a central difference to estimate \(F'(60)\text{.}\) Use this estimate as needed in subsequent questions.

  2. Find the local linearization \(y = L(t)\) to the function \(y = F(t)\) at the point where \(a = 60\text{.}\)

  3. Determine an estimate for \(F(63)\) by employing the local linearization.

  4. Do you think your estimate in (c) is too large or too small? Why?

Answer

  1. \(F'(60)\approx 1.56\) degrees per minute.

  2. \(L(t) \approx 1.56(t-60)+324.5\text{.}\)

  3. \(F(63)\approx L(63)\approx = 329.18\) degrees F.

  4. Overestimate.

Solution

  1. The central difference approximation with \(h=15\) gives

    \begin{equation*} F'(60)\approx \frac{F(75)-F(45)}{30}=\frac{342.8-296}{30}=1.56 \end{equation*}

    degrees per minute.

  2. The local linearization at \(a=60\) is \(L(t)=F'(60)(t-60)+F(60)\approx 1.56(t-60)+324.5\text{.}\)

  3. Using the linearization from (b), \(F(63)\approx L(63)\approx 1.56(3)+324.5=329.18\) degrees F.

  4. Since the values of \(F(t)\) given in the table appear to be increasing at a decreasing rate, it follows that the graph of \(F\) is concave down. This implies that near \(t=60\text{,}\) the graph of \(L(t)\) (the tangent line to \(F\) at \(t=60)\) lies above the graph of \(F\text{.}\) Therefore, the approximation given in part (c) is probably an overestimate.

7.

An object moving along a straight line path has a differentiable position function \(y = s(t)\text{;}\) \(s(t)\) measures the object's position relative to the origin at time \(t\text{.}\) It is known that at time \(t = 9\) seconds, the object's position is \(s(9) = 4\) feet (i.e., 4 feet to the right of the origin). Furthermore, the object's instantaneous velocity at \(t = 9\) is \(-1.2\) feet per second, and its acceleration at the same instant is \(0.08\) feet per second per second.

  1. Use local linearity to estimate the position of the object at \(t = 9.34\text{.}\)

  2. Is your estimate likely too large or too small? Why?

  3. In everyday language, describe the behavior of the moving object at \(t = 9\text{.}\) Is it moving toward the origin or away from it? Is its velocity increasing or decreasing?

Answer

  1. \(s(9.34) \approx L(9.34) = 3.592\text{.}\)

  2. underestimate.

  3. The object is slowing down as it moves toward toward its starting position at \(t=4\text{.}\)

Solution

  1. The linearization \(L\) of \(s\) at \(t=a\) is \(L(t) = s(a) + s'(a)(t-a)\text{.}\) With the information we have we can find the linearization of \(s\) at \(t=9\text{,}\) using the facts that \(s(9) = 4\) and \(s'(9) = -1.2\text{.}\) Thus, the linearization of \(s\) at \(t=9\) is \(L(t) = 4 - 1.2(t-9)\text{.}\) One use of the linearization is that for \(t\) close to \(9\text{,}\) \(L(t)\) can provide a good approximation to \(s(t)\text{.}\) So

    \begin{equation*} s(9.34) \approx L(9.34) = 4-1.2(0.34) = 3.592\text{.} \end{equation*}

  2. Since \(s''(4) = 0.08\) is positive, we know that the graph of \(s\) is concave up at \(t=4\text{.}\) This means that the linearization of \(s\) at \(t=4\) lies below the graph of \(s\) and so \(L(9.34)\) is an underestimate to \(s(9.34)\text{.}\)

  3. Given that \(s'(4)\) is negative, the object is moving back toward its original position. The fact that \(s''(4)\) is positive means that the velocity of the object is increasing at \(t=4\text{.}\) This means that the object is actually slowing down as it moves toward toward its starting position at \(t=4\text{.}\)

8.

For a certain function \(f\text{,}\) its derivative is known to be \(f'(x) = (x-1)e^{-x^2}\text{.}\) Note that you do not know a formula for \(y = f(x)\text{.}\)

  1. At what \(x\)-value(s) is \(f'(x) = 0\text{?}\) Justify your answer algebraically, but include a graph of \(f'\) to support your conclusion.

  2. Reasoning graphically, for what intervals of \(x\)-values is \(f''(x) \gt 0\text{?}\) What does this tell you about the behavior of the original function \(f\text{?}\) Explain.

  3. Assuming that \(f(2) = -3\text{,}\) estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\text{.}\) Is your estimate larger or smaller than the true value of \(f(1.88)\text{?}\) Justify your answer.

Answer

  1. \(x=1\text{.}\)

  2. On \(-0.37 \lt x \lt 1.37\text{;}\) \(f\) is concave up.

  3. \(f(1.88) \approx 0.02479\text{,}\) and this estimate is larger than the true value of \(f(1.88)\text{.}\)

Solution

  1. Since \(e\) raised to any power is always positive (and thus never \(0\)), it follows that the only value that makes \(f'(x) = (x-1)e^{-x^2} = 0\) is \(x=1\text{.}\) This is confirmed by plotting \(f'\) where we only see the graph cross the \(x\)-axis at \(x=1\) (though we do see the graph approach the \(x\)-axis) as \(x\) increases or decreases without bound).

  2. We know that \(f''(x) \gt 0\) wherever \(f'\) is increasing. From the graph of \(f'\text{,}\) we see that \(f'\) is increasing approximately on the interval \(-0.37 \lt x \lt 1.37\text{,}\) and thus \(f\) is concave up on this interval.

  3. The tangent line approximation at \(a = 2\) is \(L(x) = f(2) + f'(2)(x-2) \approx 0.0182 + -0.05495(x-2)\text{.}\) Thus, \(f(1.88) \approx L(1.88) \approx 0.0182 + -0.05495(1.88-2) = 0.02479\text{.}\) Since \(f''(x) \lt 0\) at and around \(x = 2\text{,}\) \(f\) is concave down there, and therefore the tangent line lies above the curve. This makes our estimate larger than the true value of \(f(1.88)\text{.}\)

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