Question

the graph of the function f(x)=x^3 - 27x has one local maximum and one local minimum point. find these points using the first - derivative test. find the first derivative of the function. f(x)=3x^2 - 27 the function has a local maximum at the point (type an ordered pair.) the function has a local minimum at the point (type an ordered pair.)

Explanation:

Step1: Find critical points

Set $f'(x)=0$, so $3x^{2}-27 = 0$. Then $3x^{2}=27$, $x^{2}=9$, and $x=\pm3$.

Step2: Test intervals

Choose test - points in the intervals $(-\infty,- 3)$, $(-3,3)$ and $(3,\infty)$. Let's choose $x=-4$, $x = 0$ and $x = 4$.
For $x=-4$, $f'(-4)=3\times(-4)^{2}-27=3\times16 - 27=48 - 27 = 21>0$.
For $x = 0$, $f'(0)=3\times0^{2}-27=-27<0$.
For $x = 4$, $f'(4)=3\times4^{2}-27=3\times16 - 27=48 - 27 = 21>0$.
Since $f'(x)$ changes sign from positive to negative at $x=-3$, the function has a local maximum at $x=-3$.
$f(-3)=(-3)^{3}-27\times(-3)=-27 + 81=54$.
Since $f'(x)$ changes sign from negative to positive at $x = 3$, the function has a local minimum at $x = 3$.
$f(3)=3^{3}-27\times3=27-81=-54$.

Answer:

The function has a local maximum at the point $(-3,54)$.
The function has a local minimum at the point $(3,-54)$.