for f(x)=2x³ + 3x² - 36x + 5, use the second derivative test to determine the local maximum of f. as your…

for f(x)=2x³ + 3x² - 36x + 5, use the second derivative test to determine the local maximum of f. as your answer, input the value of f at the point of the local maximum.

for f(x)=2x³ + 3x² - 36x + 5, use the second derivative test to determine the local maximum of f. as your answer, input the value of f at the point of the local maximum.

Answer

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=2x^{3}+3x^{2}-36x + 5$ using the power rule. The power rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $f^\prime(x)=6x^{2}+6x-36$

Step2: Set the first - derivative equal to zero

$6x^{2}+6x - 36=0$. Divide through by 6: $x^{2}+x - 6=0$. Factor the quadratic equation: $(x + 3)(x - 2)=0$. Solve for $x$: $x=-3$ or $x = 2$.

Step3: Find the second - derivative

Differentiate $f^\prime(x)=6x^{2}+6x-36$ with respect to $x$. $f^{\prime\prime}(x)=12x+6$

Step4: Evaluate the second - derivative at the critical points

For $x=-3$: $f^{\prime\prime}(-3)=12(-3)+6=-36 + 6=-30<0$. For $x = 2$: $f^{\prime\prime}(2)=12(2)+6=24 + 6=30>0$. Since $f^{\prime\prime}(-3)<0$, the function $f(x)$ has a local maximum at $x=-3$.

Step5: Find the value of the function at the local maximum

Substitute $x=-3$ into $f(x)$: $f(-3)=2(-3)^{3}+3(-3)^{2}-36(-3)+5$ $=2(-27)+3(9)+108 + 5$ $=-54 + 27+108 + 5$ $=-27+108 + 5$ $=81 + 5$ $=86$

Answer:

86