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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Analytical

1st and 2nd Derivative Tests

Maximum and Minimum on Closed IntervalsPrevious Section
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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
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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.