Calculus BC ยท Unit 5
Read the sign of f'' for concavity, inflection points, and local extrema.
The sign of f'' is the concavity
\(f''(x) > 0\) means \(f\) is concave up. \(f''(x) < 0\) means \(f\) is concave down.
Concave up is the same statement as \(f'\) increasing.
An inflection point is a concavity change
The concavity has to change. \(f''(c) = 0\) is not enough by itself. \(f''\) may also be undefined at an inflection point.
The second derivative test
If \(f'(c) = 0\) and \(f''(c) > 0\), then \(f\) has a local minimum at \(c\).
If \(f'(c) = 0\) and \(f''(c) < 0\), then \(f\) has a local maximum at \(c\).
If \(f''(c) = 0\), the test is inconclusive.
Go back to the first derivative test when the second is silent
For \(f(x) = x^4\), \(f'(0) = 0\) and \(f''(0) = 0\). The first derivative changes from negative to positive, so the point is a local minimum.
Read f'' and name the concavity.
\(f'(x) = 3x^2 - 6x\)
\(f''(x) = 6x - 6 = 6(x - 1)\)
\(f'' < 0\) on \((-\infty, 1)\), so f is concave down there.
\(f'' > 0\) on \((1, \infty)\), so f is concave up there.
\(f(1) = 1 - 3 = -2\)
Inflection point \((1, -2)\).
Answer: concave down on \((-\infty,1)\); concave up on \((1,\infty)\); inflection \((1,-2)\)
\(f''(x) = 2 > 0\)
f is concave up on \((-\infty, \infty)\). There is no inflection point.
Answer: always concave up
\(f''(x) = -2 < 0\)
f is concave down on \((-\infty, \infty)\).
Answer: always concave down
\(f'(x) = 3x^2\) and \(f'(0) = 0\)
\(f''(x) = 6x\) and \(f''(0) = 0\)
The second derivative test is inconclusive.
\(f'\) does not change sign, so \(x = 0\) is not a local extremum. It is an inflection point.
Answer: test inconclusive; inflection at 0
\(f''(x) > 0\) when \(x > 3\).
Answer: \((3, \infty)\)
\(f''(x) = e^x > 0\)
f is concave up on \((-\infty, \infty)\).
Answer: always concave up
Use f'' both for concavity and for the second derivative test.
\(f''(x) = 12x^2 - 12 = 12(x^2 - 1)\)
\(f'' = 0\) at \(x = \pm 1\)
\(f'' > 0\) on \((-\infty, -1)\) and \((1, \infty)\).
\(f'' < 0\) on \((-1, 1)\).
Concavity changes at both zeros.
\(f(1) = 1 - 6 = -5\) and \(f(-1) = -5\)
Answer: inflection points \((-1, -5)\) and \((1, -5)\)
\(f'(x) = 4x^3 - 12x = 4x(x^2 - 3)\)
Critical points: \(x = 0, \pm\sqrt{3}\)
\(f''(x) = 12(x^2 - 1)\)
\(f''(0) = -12 < 0\): local maximum.
\(f''(\sqrt{3}) = 12(3 - 1) > 0\): local minimum. The same holds at \(-\sqrt{3}\).
Answer: local max at 0; local mins at \(\pm\sqrt{3}\)
\(f''(x) = -\sin x\)
Concave down on \((0, \pi)\), where \(\sin x > 0\).
Concave up on \((\pi, 2\pi)\).
Inflection at \(x = \pi\), and \(f(\pi) = 0\).
Answer: inflection \((\pi, 0)\)
\(f'(x) = e^{x}(x + 1)\)
\(f''(x) = e^{x}(x + 2)\)
Concavity changes at \(x = -2\) from down to up.
\(f(-2) = -2e^{-2}\)
At the critical point \(x = -1\), \(f''(-1) = e^{-1} > 0\), so it is a local minimum.
Answer: inflection \(\left(-2, -2e^{-2}\right)\); local min at \(x = -1\)
\(f'(0) = 0\) and \(f''(0) = 0\)
The second derivative test is inconclusive.
\(f'(x) = 4x^3\) changes from negative to positive, so \(x = 0\) is a local minimum.
Answer: inconclusive; local min by the first derivative test
\(f''(x) = 6(x - 1) < 0\) on \((-\infty, 1)\).
Answer: A. \((-\infty, 1)\)
A quotient for f'', an undefined second derivative, and a sentence a reader can check.
\(f'(x) = \dfrac{2x}{x^2 + 1}\)
\(f''(x) = \dfrac{2(1 - x^2)}{(x^2 + 1)^2}\)
The denominator is always positive.
\(f'' > 0\) on \((-1, 1)\). Concave up.
\(f'' < 0\) on \((-\infty, -1)\) and \((1, \infty)\). Concave down.
\(f(1) = f(-1) = \ln 2\)
Answer: inflection points \((-1, \ln 2)\) and \((1, \ln 2)\)
\(f''(x) = -\dfrac{2}{9x^{5/3}}\)
For \(x > 0\), \(f'' < 0\). Concave down.
For \(x < 0\), \(x^{5/3} < 0\), so \(f'' > 0\). Concave up.
Concavity changes at \(x = 0\), where \(f''\) is undefined. Inflection point \((0, 0)\).
Answer: concave up on \((-\infty,0)\); concave down on \((0,\infty)\); inflection \((0,0)\)
\(f'(x) = 3x(x - 2)\) and \(f'(0) = 0\)
\(f''(x) = 6(x - 1)\) and \(f''(0) = -6 < 0\)
By the second derivative test, \(f\) has a local maximum at \(x = 0\).
\(f(0) = 0\)
Answer: local max \((0, 0)\)
\(f'(2) = 0\)
\(f''(2) = 6 > 0\)
Local minimum.
\(f(2) = 8 - 12 = -4\)
Answer: local min \((2, -4)\)
\(f'\) increasing means \(f\) is concave up.
\(f'\) decreasing means \(f\) is concave down.
f is concave up on \((-2, 1)\) and concave down on \((1, 4)\).
f has an inflection point at \(x = 1\).
Answer: concave up then down; inflection at \(x = 1\)
\(f''(0) = 0\)
The first derivative changes from negative to positive.
Answer: A. inconclusive, and it is a local minimum
Need help with concavity?
Book a Session with Amy