Use calculus and a graphing calculator to find the approximate location of all relative extrema.
-$f(x)=x^{5}-15 x^{4}-3 x^{3}-172 x^{2}+135 x+0.062$
A) Approximate local maximum at 0.379 ; approximate local minimum at 12.565
B) Approximate local maximum at 0.4 ; approximate local minimum at -12.534
C) Approximate local maximum at 0.379 ; approximate local minima at -0.472 and 12.565
D) Approximate local maximum at 0.478 ; approximate local minima at -0.55 and -12.5

Question

Quizplus · Accepted Answer

The correct answer is A because the first derivative of the function is $f'(x)=5x^4-60x^3-9x^2-344x+135$, and the second derivative is $f''(x)=20x^3-180x^2-18x-344$. The first derivative is zero at $x \approx 0.379$ and $x \approx 12.565$, and the second derivative is negative at $x \approx 0.379$ and positive at $x \approx 12.565$, so there is a local maximum at $x \approx 0.379$ and a local minimum at $x \approx 12.565$.

Use Calculus and a Graphing Calculator to Find the Approximate f(x)=x5−15x4−3x3−172x2+135x+0.062f(x)=x^{5}-15 x^{4}-3 x^{3}-172 x^{2}+135 x+0.062f(x)=x5−15x4−3x3−172x2+135x+0.062

Use Calculus and a Graphing Calculator to Find the Approximate $f(x)=x^{5}-15 x^{4}-3 x^{3}-172 x^{2}+135 x+0.062$