CUET Preparation Today
The point at which the function $'f'$ given by $f(x)=(x-2)^4(x+1)^3$ has point of inflexion is : |
$x=\frac{2}{7}$ $x=2$ $x=-1$ $x=-2$ |
$x=-1$ |
The correct answer is Option (3) → $x=-1$ $f(x)=(x-2)^4(x+1)^3$ $f'(x)=4(x-2)^3(x+1)^3+3(x-2)^4(x+1)^2$ $=(x-2)^3(x+1)^2(7x-2)$ $f''(x)=(x-2)^2(x+1)(7x-2)(12x-15)$ for point of inflection, $f''(x)=0$ $(x+1)=0⇒x=-1$ |
