How do you find local maximum value of f using the first and second derivative tests: #2x^3-3x^2-36x-3#?
1 Answer
There is a local maximum of
Explanation:
1. Find the critical values of the function. These are where the local maximum could exist.
A function's critical values exist where the function's derivative equals
To find this function's derivative, use the power rule.
#f(x)=2x^3-3x^2-36x-3#
#f'(x)=6x^2-6x-36#
Now, we should find the critical values by setting
#6x^2-6x-36=0#
Divide the terms by
#x^2-x-6=0#
#(x-3)(x+2)=0#
#x=3" "# or#" "x=-2#
These are the function's two critical values. These are where the local maxima or minima could exist.
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2. Use the first derivative test to see if the function switches from increasing to decreasing or decreasing to increasing around the critical values.
Test the sign of the derivative around the points at
- If
#f'(x)# switches from increasing to decreasing around a critical value, there is a local maximum at the critical value. - If
#f'(x)# switches from decreasing to increasing around a critical value, there is a local minimum at the critical value.
#f'(color(red)(-3))=6(-3)^2-6(-3)-36=36larrcolor(blue)("INCREASING")#
#f'(color(red)(-2))=6(-2)^2-6(-2)-36=0#
#f'(color(red)(-1))=6(-1)^2-6(-1)-36=-24larrcolor(blue)("DECREASING")#
Since the derivative changes from increasing to decreasing at
Since we can easily envision the graph of a cubic, we know that there should be a local minimum at
#f'(color(red)2)=6(2)^2-6(2)-36=-24larrcolor(blue)("DECREASING")#
#f'(color(red)3)=6(3)^2-6(3)-36=0#
#f'(color(red)4)=6(4)^2-6(4)-36=36larrcolor(blue)("INCREASING")#
As we suspected it would, the derivative changes from decreasing to increasing, revealing a local minimum of
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3. Examine the concavity at each critical value through the second derivative (the second derivative test).
First, find the second derivative by differentiating the first derivative.
#f'(x)=6x^2-6x-36#
#f''(x)=12x-6#
The process for the second derivative test is as follows:
- If the value of the second derivative at a critical value is positive, the function is concave up at that point and there is a local minimum.
- If the value of the second derivative at a critical value is negative, the function is concave down at that point and there is a local maximum.
Note that we should get the same results that the first derivative test gave us--this is just another way of reaching the same conclusion.
#f''(color(red)(-2))=12(-2)-6=-30larrcolor(blue)("CONCAVE DOWN")#
Thus, there is a local maximum at
#f''(color(red)3)=12(3)-6=30larrcolor(blue)("CONCAVE UP")#
Thus, there is a local minimum at
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If you're able, always consult a graph of the function to check your answer:
graph{2x^3-3x^2-36x-3 [-5, 7, -106.8, 107]}