Answer to Question #145769 in Calculus for liam donohue

Question

Suppose that it is given to you that
f′(x)=(x+6)(12−x)(x−15)

Then the first relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative
?
at this point.

The second relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative
?
at this point.

The third relative extremum (from the left) for f(x) occurs at x=

The function f(x) has a relative
?
at this point.

The first inflection point (from the left) for f(x) occurs at x=


The second inflection point (from the left) for f(x) occurs at x=

Answer 1

Given Derivative is

$f'(x)=(x+6)(12-x)(x-15)$

For calculating critical points putting $f'(x)=0$

critical points are $-6,12,15$ respectively.

The possible intervals are $(-\infty,-6),(-6,12),(12,15) ,(15,\infin)$

Sign of $f'(x) in (-\infty,-6)$ is $(+)$ $\implies f'(x)>0\implies f(x)$ is increasing.

Sign of $f'(x) in (-6,12)$ is $(-)$ $\implies f'(x)<0\implies f(x)$ is Decreasing.

Sign of $f'(x) in (12,15)$ is $(+)$ $\implies f'(x)>0\implies f(x)$ is increasing.

Sign of $f'(x) in (15,\infty)$ is $(-)$ $\implies f'(x)<0\implies f(x)$ is decreasing.

$\Rightarrow$ The first relative extremum (from the left) for $f(x)$ occurs at $x=-6$

The function $f(x)$ has a relative maximum at this point.

$\Rightarrow$ The second relative extremum (from the left) for $f(x)$ occurs at $x=12$

The function $f(x)$ has a relative minimum at this point.

$\Rightarrow$ The third relative extremum (from the left) for $f(x)$ occurs at $x=15$

The function $f(x)$ has a relative maximum at this point.

There is no inflexion point for the given derivative.

Since the function changes its value for every interval.