Question

A cubic function of x has maximum value 10 and minimum $\frac{-5}{2}$ when $x=-3,x=2$ then the function is

• A. $\frac{1}{5}{x}^3+\frac{3}{10}{x}^2-\frac{18}{5}x+\frac{19}{10}$
• B. $x^3+3x^2-18x+19$
• C. $2x^3+3x^2-36x+10$
• D. $x^3+x^2+x+1$

Answer 1

The correct option is A $\frac{1}{5}{x}^3+\frac{3}{10}{x}^2-\frac{18}{5}x+\frac{19}{10}$

Given $f^{\prime }(-3)=0,f^{\prime }\left(2\right)=0$
Also, $f(-3)=10$ and $f\left(2\right)=-\frac{5}{2}$
Now, we will check by options.
For option A, $f^{\prime }\left(x\right)=\frac{6(x^2+x-6)}{10}$
$\Rightarrow x=-3,2$ are the critical values.
$f^{\prime \prime }\left(x\right)=\frac{3(2x+1)}{5}$
$f^{\prime \prime }(-3)=-3<0$ and $f^{\prime \prime }\left(2\right)=3>0$
Hence, f has a maximum at $x=-3$ and minimum at $x=2$.
$f(-3)=10$ and $f\left(2\right)=-\frac{5}{2}$
Option A matches with all given conditions.
Hence $f\left(x\right)=\frac{1}{5}{x}^3+\frac{3}{10}{x}^2-\frac{18}{5}x+\frac{19}{10}$