Question

The slope of f(x)=x³−12x²+36x−48 is zero at which value/s of x?

Answer 1

$$y'=(x^3-12x^2+36x-48)'$$

$$=3x^2-12(2x)+36(1)-0$$

$$=3x^2-24x+36$$

If the slope is zero 

$$3x^2-24x+36=0$$

$$x^2-8x+12=0$$

$$(x-2)(x-6)=0$$

$$x_1=2, x_2=6$$

The slope of $f(x)=x^3−12x^2+36x−48$ is zero at $x=2$ or at $x=6.$