Question

If $f(x)=3x^3-9x^2-27x+15$, then the maximum value of $f(x)$ is .......

• A. $-66$
• B. $30$
• C. $-30$
• D. $66$

Answer 1

Answer: (B)

We have,
$f(x)=3x^3-9x^2-27x+15$
$\Rightarrow f^{\prime }(x)=9x^2-18x-27$
For maxima or minima, we put $f^{\prime }(x)=0$
$\Rightarrow 9x^2-18x-27=0$
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0\Rightarrow x=-1,3$
Now, $f^{\prime \prime }(x)=18x-18$
at $x=-1,f^{\prime \prime }(x)=18(-1)-18=-36<0$
So, $x=-1,$ point of maxima
$\therefore$ Maximum value of $f(x)$ at $x=-1$ is
$f(-1)=3(-1{)}^3-9(-1{)}^2-27(-1)+15$
$=-3-9+27+15=-12+42=30$