Question

Let $f\left(x\right)=\cos \pi x+10x+3x^2+x^3,-2\le x\le 3.$ The absolute minimum value of $f\left(x\right)$ is

• A. $0$
• B. $-15$
• C. $3-2\pi$
• D. none of these

Answer 1

The correct option is B $-15$

$f\left(x\right)=cos\pi x+10x+3x^2+x^3$
$f^{\prime }\left(x\right)=3x^2+6x+10-\pi sin\pi x=3(x+1{)}^2+7-\pi \sin \pi x$
since $sin\pi x\le 1\Rightarrow 7-\pi \sin \pi x>0$
Hence $f^{\prime }\left(x\right)>0$
Ergo $f\left(x\right)$ is strictly increasing
$\Rightarrow f_{min}=f(-2)=cos(-2\pi )-20+12-8=-15$