Question

Find the maximum and minimum values of $f\left(x\right)=2x^3-3x^2-12x+15$ on the closed interval $[0,3].$

• A. $f_{max}=15$
• B. $f_{max}=10$
• C. $f_{min}=5$
• D. $f_{min}=-5$

Answer 1

Answer: (A), (D)

$f_{max}=15$
$f_{min}=-5$

The derivative of $f$ is $f^{\prime }\left(x\right)=6x^2-6x-12=6(x-2)(x+1)$

So the critical points of $f$ are the solution of the equation $6(x-2)(x+1)=0$

and the numbers $c$ for which $f^{\prime }\left(c\right)$ does not exits.

There are none of the latter, so the critical of $f$ occur at $x=-1$ and $x=2$.

the first of these is not in the domain of $f;$

We discard it, and thus the only critical point of $f$ in $[0,3]$ is $x=2$.

Including the two endpoints, our list of all values of $x$ that yeilds a possible maximum or minimum value of $f$ consists of $0,2,$ and $3.$

We evaluate the function $f$ at each:

$f\left(0\right)=15,f\left(2\right)=-5)$ and $f\left(3\right)=6$.

Therefore the maximum value of $f$ on $[0,3]$ is $f\left(0\right)=15$ and its minimum value if $f\left(2\right)=-5.$