Question

The- maximum value of function $x^3-12x^2+36x+17$ in the interval $[1,10]$ is

• A. $17$
• B. $177$
• C. $77$
• D. None of these

Answer 1

Answer: (B)

Let $f(x)=x^3-12x^2+36x+17$
$\therefore f^{\prime }(x)=3x^2-24x+36=0$
For maxima, put $f^{\prime }(x)=0$
$\Rightarrow 3x^2-24x+36=0$
$\Rightarrow (x-2)(x-6)=0$
$\Rightarrow x=2,6$
Again, $f^{\prime \prime }(x)=6x-24$ is negative at $x=2$
So that, $f(6)=17,f(2)=49$
At the end points, $f(1)=42,f(10)=177$
So that, $f(x)$ has its maximum value $177$ .