CUET Preparation Today
The slope of the function $f(x)=-\frac{x^3}{3}+\frac{5x^2}{2}-6x+12 $ is maximum at x= __________. |
6 -1 $\frac{5}{2}$ -2 |
$\frac{5}{2}$ |
The correct answer is Option (3) → $\frac{5}{2}$ $f(x)=-\frac{x^3}{3}+\frac{5x^2}{2}-6x+12$ Slope = $f'(x)$ $=-x^2+5x-6$ and, thus critical points is, $f''(c)=0$ $⇒-2x+5=0$ $⇒x=\frac{5}{2}$ |
