High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 1 Textbook > Applications of Differentiation

1st and 2nd Derivative Tests

Critical Points and ExtremaPrevious Section
Maximum and Minimum on Closed IntervalsNext Section
Popular Courses
Find My Course
0:00 / 0:00

The 1st and 2nd Derivative Tests

When a functions switches from increasing to decreasing (or vise versa) there will be a local maximum (or minimum) value. Often we'd like to know what type of extrema occurs at critical points. We can classify our critical points using the following derivative tests.

First Derivative Test

Suppose f(x)f\left(x\right)is a continuous function on [a,b]\left[a,b\right], differentiable on (a,b)\left(a,b\right) , and PP in (a,b)\left(a,b\right) is a critical point of f(x)f\left(x\right)
  • PPis a relative minimum of f(x)f\left(x\right)if f′(x)f'\left(x\right) changes from negative to positive at PP
  • PPis a relative maximum of f(x)f\left(x\right)if f′(x)f'\left(x\right)changes from positive to negative at PP

Second Derivative Test

Assume f(x)f\left(x\right) is a function such that f′′(x)f''\left(x\right)is continuous on an open interval containing the point PP
  • if f′(P)=0f'\left(P\right)=0 and f′′(P)>0f''\left(P\right)>0 , then PPis a relative minimum
  • if f′(P)=0f'\left(P\right)=0 and f′′(P)<0f''\left(P\right)<0 , then PPis a relative maximum
  • if f′(P)=0f'\left(P\right)=0 and f′′(P)=0f''\left(P\right)=0 , then the test is inconclusive
0:00 / 0:00

Example: Finding Relative Extrema

Find the relative extrema (max or min) of the function f(x)=ex(x2−3)+5f\left(x\right)=e^x\left(x^2-3\right)+5

Using the First Derivative Test:
1. Find the first derivative
f′(x)=ex(x2−3)+ex(2x)=ex(x2+2x−3)f'\left(x\right)=e^x\left(x^2-3\right)+e^x\left(2x\right)=e^x\left(x^2+2x-3\right)

2. Determine when the first derivative is 0 or undefined -- these are the critical points
f′(x)=0f'\left(x\right)=0

ex(x2+2x−3)=0e^x\left(x^2+2x-3\right)=0

Since exe^x never equals 0, we need x2+2x−3=0 → (x−1)(x+3)=0x^2+2x-3=0\ \to\ \left(x-1\right)\left(x+3\right)=0
So, the critical points are when x=−3  or  x=1x=-3\ \text{ or }\ x=1

3. Determine the intervals of increasing/decreasing


4. Find the y value that corresponds to that critical point
f(1)=e1(12−3)+5=5−2ef\left(1\right)=e^1\left(1^2-3\right)+5=5-2e

f(−3)=e−3((−3)2−3)+5=5+6e−3f\left(-3\right)=e^{-3}\left(\left(-3\right)^2-3\right)+5=5+6e^{-3}

5. Make a conclusion
Since the graph is increasing to the left of x=−3x=-3 and decreasing to the right of x=−3x=-3, (−3,5+6e−3)(-3, 5+6e^{-3}) is a relative maximum.
Since the graph is decreasing to the left of x=1x=1 and increasing to the right of x=1x=1, (1,5−2e)(1,5-2e) is a relative minimum.


Using the Second Derivative Test:
1. Find the first and second derivatives
f′(x)=ex(x2−3)+ex(2x)=ex(x2+2x−3)f'\left(x\right)=e^x\left(x^2-3\right)+e^x\left(2x\right)=e^x\left(x^2+2x-3\right)

f′′(x)=ex(x2+2x−3)+ex(2x+2)=ex(x2+4x−1)f''\left(x\right)=e^x\left(x^2+2x-3\right)+e^x\left(2x+2\right)=e^x\left(x^2+4x-1\right)

2. Determine when the first derivative is 0 or undefined -- these are the critical points
f′(x)=0f'\left(x\right)=0

ex(x2+2x−3)=0e^x\left(x^2+2x-3\right)=0

Since exe^x never equals 0, we need x2+2x−3=0 → (x−1)(x+3)=0x^2+2x-3=0\ \to\ \left(x-1\right)\left(x+3\right)=0
So, the critical points are when x=1  or  x=−3x=1\ \text{ or }\ x=-3

3. Substitute each critical point into the second derivative
f′′(1)=e1(12+4(1)−1)=4e>0   →  x=1f''\left(1\right)=e^1\left(1^2+4\left(1\right)-1\right)=4e>0\ \ \ \to\ \ x=1 corresponds to a local min

f′′(−3)=e−3((−3)2+4(−3)−1)=−4e−3<0   →   x=−3f''\left(-3\right)=e^{-3}\left(\left(-3\right)^2+4\left(-3\right)-1\right)=-4e^{-3}<0\ \ \ \to\ \ \ x=-3 corresponds to a local max

4. Find the y value that corresponds to that critical point
f(1)=e1(12−3)+5=5−2ef\left(1\right)=e^1\left(1^2-3\right)+5=5-2e

f(−3)=e−3((−3)2−3)+5=5+6e−3f\left(-3\right)=e^{-3}\left(\left(-3\right)^2-3\right)+5=5+6e^{-3}

Therefore, the relative min is(1, 5−2e)\left(1,\ 5-2e\right) and the relative max is (−3, 5+6e−3)\left(-3,\ 5+6e^{-3}\right)

Example: 2nd Derivative Test

Use the second derivative test to classify the relative extrema of f(x)=−3x5+5x3f(x)=-3x^{5}+5x^{3}.
f′(x)=−15x4+15x2f′(x)=−15x2(x2−1)⇒f′(x)=0  when  x=−1,0,1f′′(x)=−60x3+30xNow, we need to check the concavity at each of our critical pointsf′′(−1)=−60(−1)3+30(−1)=60−30=30>0f′′(0)=−60(0)3+30(0)=0f′′(1)=−60(1)3+30(1)=−60+30=−30<0So, x=−1 is a local minimum, because f is concave up at that point,The test is inconclusive at x=0, because f has an inflection point there,and, x=1 is a local maximum because f is concave down at that point.\begin{aligned}

f'(x) =& -15x^4 +15x^2 \\
f'(x) =& -15x^2(x^2-1) \\
\Rightarrow f'(x) =&0 \  \text{ when }  \ x=-1,0,1 \\
\text{} \\
f''(x)=& -60x^3 +30x\\
\text{Now, we }& \text{need to check the concavity at each of our critical points} \\
f''(-1) =& -60(-1)^3 + 30(-1) = 60-30 = 30 >0 \\
f''(0) =& -60(0)^3 + 30(0) = 0 \\
f''(1) =& -60(1)^3 + 30(1) = -60+30 = -30 <0 \\
\text{} \\
\text{So, } x=&-1 \text{ is a local minimum, because } f\text{ is concave up at that point,} \\
\text{The test} &\text{ is inconclusive at } x=0 \text{, because } f \text{ has an inflection point there,} \\
\text{and, } x=&1  \text{ is a local maximum because } f\text{ is concave down at that point.}
\end{aligned}
Find and classify the critical points of f(x)=xex−exf(x)=xe^{x}-e^{x} as relative maximums or minimums.
Extra Practice
Critical points and extrema
Find all local extreme values of the function f(x)=x3−6x2+9x+1f(x)=x^3-6x^2+9x+1
Derivative tests
Let f(x)=−x3+x2+xf(x) = -x^3+x^2+x and use the first derivative test to find and classify the extreme values.
Since f is a negative cubic and we’re not given an interval, f has no absolute max or min. Here, f’(x)=−3x2+2x+1f’(x)=-3x^2+2x+1, which exists everywhere. Let’s set it equal to zero and solve for x:
0=−3x2+2x+1=−(3x+1)(x−1)→x=1,−1/3.0 = -3x^2+2x+1 = -(3x+1)(x-1) \rightarrow x=1,-1/3.
Critical points and Extrema
Given this table of values, which of the following statements must NOT be true?
f(x)f′(x)f′′(x)x=0313x=100−1x=20−20\begin{array}{|c|c|c|c|} \hline &f(x)&f'(x)&f''(x)\\ \hline x=0&3&1&3\\ x=1&0&0&-1\\ x=2&0&-2&0\\ \hline \end{array}
Absolute Extrema on Closed Interval
Find the absolute maximum of f(x)=3x4−4x3\displaystyle f(x)=3x^{4}-4x^{3} on the interval [−1,2][-1,2] using the second derivative test.
Absolute Extrema on Closed Interval
Find the absolute extrema of f(x)=3x4−4x3\displaystyle f(x)=3x^{4}-4x^{3} on the interval [−1,2][-1,2] using the second derivative test.
Relative Extrema
Find and classify all local and absolute extrema of f(x)=sin⁡xcos⁡xf\left(x\right)=\sin x\cos x on the interval [0,  2π]\left[0,\ \ 2\pi\right].
Critical Points and Extrema: Max and Min
Find and classify the critical points of f(x)=xex−exf(x)=xe^{x}-e^{x} as relative maxima or minima.
Practice: Absolute Extrema on Closed Interval
Q.\textbf{Q.} Find the absolute maximum and minimum values of f(x)=6x43−3x13\displaystyle f(x)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}} on the interval [−1,1][-1,1] and then everywhere, using the second derivative test.
Absolute Extrema on Closed Interval
Find the absolute extrema of f(x)=3x4−4x3\displaystyle f(x)=3x^{4}-4x^{3} on the interval [−1,2][-1,2] using the second derivative test.
Continuity
Let f(x)f(x) be a continuous function on the open interval (a,b)(a, b). Which of the following four statements are always true.
Select all that apply.
Derivative Tests
Which of the following five functions are concave up on their whole domain? Select all that apply.
Critical points and Extrema
Given this table of values, which of the following statements must NOT be true?
f(x)f′(x)f′′(x)x=0313x=100−1x=20−20\begin{array}{|c|c|c|c|} \hline &f(x)&f'(x)&f''(x)\\ \hline x=0&3&1&3\\ x=1&0&0&-1\\ x=2&0&-2&0\\ \hline \end{array}
Relative Extrema
Find and classify all local and absolute extrema of f(x)=sin⁡xcos⁡xf\left(x\right)=\sin x\cos x on the interval [0,  2π]\left[0,\ \ 2\pi\right].
Critical Points and Extrema
Let f(x)f(x) be a smooth function (a function that has infinitely many derivatives that exist everywhere) and consider x=ax = a . Which of the following statements are correct? Check off all correct statements.
Practice
f(x)=1x2+1\displaystyle f(x)=\frac{1}{x^2+1}.
The first and the second derivatives are
f′(x)=−2x(x2+1)2 and f′′(x)=6x2−2(x2+1)3.\boxed{f^{\prime}(x)=\frac{-2x}{\left(x^{2}+1\right)^{2}}\quad\text{ and }\quad f^{\prime\prime}(x)=\frac{6x^{2}-2}{\left(x^{2}+1\right)^3}.}
Practice
f(x)=1x2+1\displaystyle f(x)=\frac{1}{x^2+1}.
The first and the second derivatives are
f′(x)=−2x(x2+1)2 and f′′(x)=6x2−2(x2+1)3.\boxed{f^{\prime}(x)=\frac{-2x}{\left(x^{2}+1\right)^{2}}\quad\text{ and }\quad f^{\prime\prime}(x)=\frac{6x^{2}-2}{\left(x^{2}+1\right)^3}.}
Practice: Minimum Value (Similar to April 2018 Q45)
Find the minimum value of f(x)=14x4+13x3f(x)=\frac{1}{4}x^4+\frac{1}{3}x^3
Practice: Max, Min, and Inflection Points
Practice Question: Max, Min, and Inflection Points
Given the following table of values, which statement(s) must always be true about f(x)f\left(x\right)?
i.) f(x)f\left(x\right) has 3 critical points
Derivative tests
Let f(x)=−x3+x2+xf(x) = -x^3+x^2+x and use the first derivative test to find and classify the extreme values.
Since f is a negative cubic and we’re not given an interval, f has no absolute max or min. Here, f’(x)=−3x2+2x+1f’(x)=-3x^2+2x+1, which exists everywhere. Let’s set it equal to zero and solve for x:
0=−3x2+2x+1=−(3x+1)(x−1)→x=1,−1/3.0 = -3x^2+2x+1 = -(3x+1)(x-1) \rightarrow x=1,-1/3.
Critical points and extrema
Find all local extreme values of the function f(x)=x3−6x2+9x+1f(x)=x^3-6x^2+9x+1
Practice: Absolute Extrema on Closed Interval
Q.\textbf{Q.} Find the absolute maximum and minimum values of f(x)=6x43−3x13\displaystyle f(x)=6x^{\frac{4}{3}}-3x^{\frac{1}{3}} on the interval [−1,1][-1,1] and then everywhere, using the second derivative test.
Absolute Extrema on Closed Interval
Find the absolute extrema of f(x)=3x4−4x3\displaystyle f(x)=3x^{4}-4x^{3} on the interval [−1,2][-1,2] using the second derivative test.
Critical Points and Extrema: Max and Min
Find and classify the critical points of f(x)=xex−exf(x)=xe^{x}-e^{x} as relative maxima or minima.
Derivative Tests
Find the local extreme values of h(x)=x4−2x2h(x)=x^4-2x^2 using the second derivative test.