Predict whether the estimate is high or low
A line below the curve
The curve is \(f(x) = e^x\), and the orange tangent line at \(x = 0\) is \(L(x) = 1 + x\). The red segment at the target is the gap between the estimate and the true value.
At \(x = 0.5\) the line is at 1.5 and the curve is at about 1.65, so the estimate is too low.
At \(x = -0.5\) the line is at 0.5 and the curve is at about 0.61. The estimate is too low again, even though the line rises.
The curve bends upward, so it lies above its tangent line on both sides of the point. The bend of a curve is its concavity, and a curve that bends upward is concave up.
The bend decides
This curve is concave up, so it lies above its tangent line and the estimate is too low, an underestimate.
Press the button to bend the curve downward. It is now concave down, so it lies below the line and the estimate is too high, an overestimate.
The label in the corner shows the sign of the second derivative \(f''\), which is green when positive and red when negative. The line rises in both pictures, so its slope does not decide the answer.
Notation
Find the sign of \(f''\) between the center \(a\) and the target \(x\). The second derivative is the rate of change of the slope, so its sign tells which way the curve bends.
\[f''(x) \ \text{on the interval from } a \text{ to } x\]If \(f'' > 0\) on the whole interval, the curve is concave up. The tangent line lies below the curve, so the estimate is too low.
\[f'' > 0 \ \Rightarrow\ L(x) \le f(x)\]If \(f'' < 0\) on the whole interval, the curve is concave down. The tangent line lies above the curve, so the estimate is too high.
\[f'' < 0 \ \Rightarrow\ L(x) \ge f(x)\]Do not use the sign of \(f'\) for this decision.
The sign of \(f'\) tells only whether the line rises or falls.
When it fails
The rule needs \(f''\) to keep one sign on the whole interval. For \(f(x) = x^3\) with \(a = -0.5\), \(f'' = 6x\) changes sign at 0, so the curve bends downward to the left of 0 and upward to the right.
At \(x = 0.5\) the line is above the curve, and at \(x = 1.5\) it is below. The line crosses the curve at \(x = 1\), so neither rule decides, and you must compute both values.
Example 1: A fourth power
Use the tangent line to \(f(x) = x^4\) at \(a = 2\) to estimate \(f(2.1)\). Is the estimate too high or too low?
Find the line. The value is \(f(2) = 16\) and the slope is \(f'(2) = 4 \cdot 2^3 = 32\).
\[L(x) = 16 + 32(x - 2)\]Substitute 2.1, so the offset \(x - 2\) is 0.1.
\[L(2.1) = 16 + 3.2 = \cy{19.2}\]The second derivative is positive on the interval from 2 to 2.1, so the curve is concave up there. The label is green because \(f''\) is positive.
\[f''(x) = 12x^2 > 0 \ \text{on } [2,\,2.1]\]The tangent line lies below the curve, so the estimate is too low.
As a check, \(2.1^4 \approx 19.45\).
Example 2: The natural logarithm
Use the tangent line to \(f(x) = \ln x\) at \(a = 1\) to estimate \(\ln 1.2\). Is the estimate too high or too low?
Find the line. The value is \(f(1) = 0\) and the slope is \(f'(1) = 1\). The estimate is the line at 1.2.
\[L(x) = x - 1, \qquad L(1.2) = \cy{0.2}\]A tempting wrong move: The derivative \(1/x\) is positive, so \(f\) rises. The curve is above the line, so the estimate is too low.
In the picture, the curve lies below the orange line, and the true value, about 0.182, is less than 0.2. A positive \(f'\) shows only that the line rises.
The second derivative is negative for \(x > 0\), so the curve is concave down on the interval from 1 to 1.2. The label is red because \(f''\) is negative.
\[f''(x) = -\frac{1}{x^2} < 0\]The tangent line lies above the curve, so the estimate 0.2 is too high.
Practice
Use the tangent line to \(f(x) = \sin x\) at \(a = \pi/2\) to estimate \(\sin(\pi/2 + 0.2)\). Is the estimate too high or too low?
The tangent line at the top of the wave is horizontal. Decide which side of the line the curve is on.
Solution
The value is \(\sin(\pi/2) = 1\). The slope is \(\cos(\pi/2) = 0\), so the line is horizontal and the estimate is 1.
\[L(x) = 1\]The second derivative is \(-\sin x\). Between \(\pi/2\) and \(\pi/2 + 0.2\), \(\sin x\) is positive, so \(f''\) is negative.
\[f''(x) = -\sin x < 0\]The curve is concave down, so it lies below the line. The estimate 1 is too high.
The line is flat, so only the bend can decide.
Questions
The figure shows the curve of \(h\), with a horizontal tangent line at \(x = 3\) and, after the second answer, at \(x = 1\).
Use the tangent line at \(x = 3\) to estimate \(h(3.2)\). Is the estimate too high or too low?
- Too low
- Too high
- Correct, because the slope is 0
Answer
Too low
Use the tangent line at \(x = 1\) to estimate \(h(1.2)\). Is the estimate too high or too low?
- Too high
- Too low
- It cannot be decided, because the curve changes concavity
Answer
Too high
Use the tangent line to \(f(x) = e^{-x}\) at \(a = 0\) to estimate \(f(0.1)\). Is the estimate too high or too low?
- Too low, because \(f'' > 0\)
- Too high, because \(f' < 0\)
- Too high, because \(f'' > 0\)
Answer
Too low, because \(f'' > 0\)
Summary
You find the sign of \(f''\) between the center and the target: positive means too low, and negative means too high.
This works because a curve that bends upward lies above its tangent line, and a curve that bends downward lies below it.
You use \(f''\), not \(f'\), and you check that \(f''\) keeps its sign on the whole interval.
This is one of four skills in topic 4.6 of Which & Why: AP Calculus.