Use calculus and a graphing calculator to find the approximate location of all relative extrema.
-$f(x)=0.1 x^{4}-x^{3}-15 x^{2}+59 x+14$
A) Approximate local maximum at 1.735; approximate local minima at -6.777 and 12.542
B) Approximate local maximum at 1.755 ; approximate local minima at -6.703 and 12.552
C) Approximate local maximum at 1.661 ; approximate local minima at -6.768 and 12.52
D) Approximate local maximum at 1.791; approximate local minima at -6.785 and 12.601

Question

Quizplus · Accepted Answer

The correct answer is A because the first derivative of the function is $f'(x)=0.4x^3-3x^2-30x+59$ and the second derivative is $f''(x)=1.2x^2-6x-30$. The first derivative is zero at $x=1.735, -6.777, 12.542$, and the second derivative is negative at $x=1.735$ and positive at $x=-6.777, 12.542$. Therefore, the function has a local maximum at $x=1.735$ and local minima at $x=-6.777, 12.542$.

Use Calculus and a Graphing Calculator to Find the Approximate f(x)=0.1x4−x3−15x2+59x+14f(x)=0.1 x^{4}-x^{3}-15 x^{2}+59 x+14f(x)=0.1x4−x3−15x2+59x+14

Use Calculus and a Graphing Calculator to Find the Approximate $f(x)=0.1 x^{4}-x^{3}-15 x^{2}+59 x+14$