Use calculus and a graphing calculator to find the approximate location of all relative extrema.
-$f(x)=0.01 x^{5}-x^{4}+x^{3}+8 x^{2}-7 x-10$
A) Approximate local maxima at -1.861 and 2.247 ; approximate local minima at 0.423 and 79.192
B) Approximate local maxima at -1.861 and 2.247 ; approximate local minimum at 0.423
C) Approximate local maxima at -1.78 and 2.164 ; approximate local minima at 0.464 and 79.109
D) Approximate local maxima at -1.774 and 2.263 ; approximate local minima at 0.467 and 79.105

Question

Quizplus · Accepted Answer

The correct answer is A because the first derivative of the function is $f'(x)=0.05x^4-4x^3+3x^2+16x-7$, and the second derivative is $f''(x)=0.2x^3-12x^2+6x+16$. The first derivative is zero at $x=-1.861, 0.423, 2.247, 79.192$, and the second derivative is positive at $x=-1.861, 2.247$ and negative at $x=0.423, 79.192$. Therefore, the function has approximate local maxima at $x=-1.861, 2.247$ and approximate local minima at $x=0.423, 79.192$.

Use Calculus and a Graphing Calculator to Find the Approximate f(x)=0.01x5−x4+x3+8x2−7x−10f(x)=0.01 x^{5}-x^{4}+x^{3}+8 x^{2}-7 x-10f(x)=0.01x5−x4+x3+8x2−7x−10

Use Calculus and a Graphing Calculator to Find the Approximate $f(x)=0.01 x^{5}-x^{4}+x^{3}+8 x^{2}-7 x-10$