the graph of the function f(x)=x³ - 12x has one local maximum and one local minimum point. find these points…

Explanation:

Step1: Find the derivative

Differentiate $f(x)=x^{3}-12x$ using power - rule. The derivative $f'(x)=3x^{2}-12$.

Step2: Find critical points

Set $f'(x) = 0$. So, $3x^{2}-12 = 0$. Factor out 3: $3(x^{2}-4)=0$, then $x^{2}-4=(x - 2)(x + 2)=0$. The critical points are $x=-2$ and $x = 2$.

Step3: Use first - derivative test

Choose test points in the intervals $(-\infty,-2)$, $(-2,2)$ and $(2,\infty)$. Let's choose $x=-3$, $x = 0$ and $x=3$. For $x=-3$, $f'(-3)=3\times(-3)^{2}-12=3\times9 - 12=27-12 = 15>0$. For $x = 0$, $f'(0)=3\times0^{2}-12=-12<0$. For $x=3$, $f'(3)=3\times3^{2}-12=3\times9 - 12=15>0$. Since $f(x)$ is increasing on $(-\infty,-2)$ and decreasing on $(-2,2)$, $f(x)$ has a local maximum at $x=-2$. $f(-2)=(-2)^{3}-12\times(-2)=-8 + 24=16$. Since $f(x)$ is decreasing on $(-2,2)$ and increasing on $(2,\infty)$, $f(x)$ has a local minimum at $x = 2$. $f(2)=2^{3}-12\times2=8-24=-16$.

Answer:

The function has a local maximum at the point $(-2,16)$. The function has a local minimum at the point $(2,-16)$.