Calculate the derivative of $f$.
$f'(x)= $
The first step in locating maxima and minima is finding critical points of $f$. Since the derivative exists everywhere, the only critical points are those where $f'(x)=0$.
You should have calculated $f'(x)$ is a polynomial times an exponential function (actually, a polynomial times $f$ itself). Can the exponential function ever be zero?
. Therefore, the only critical points are where the polynomial is zero, i.e., where
$= 0$.
To find these critical points, factor out an $x$ from the polynomial (even better, factor out $-\frac{x}{6}$), so that the remaining factor is a quadratic polynomial. The factored equation becomes
$= 0$.
Now, factor the quadratic polynomial (or use the quadratic formula) to find all the critical points. Critical points: $x=$
(Separate multiple answers by commas.)
Find the intervals where $f$ is increasing and decreasing. In this case, the critical points divide the number line into four intervals. In each interval, you can test $f'(x)$ for a point in the interval to determine if $f'(x)$ is positive or negative in the interval. From left to right, list the intervals and whether $f$ is increasing or decreasing in the interval.
For the interval
, $f$ is
.
For the interval
, $f$ is
.
For the interval
, $f$ is
.
For the interval
, $f$ is
.
Local maxima can only occur where $f$ changes from increasing to decreasing (which can happen only at critical points). Find the local maxima of $f$.
$f$ has local maxima at: $x=$
and at $x=$
(Enter in increasing order.)
What values does $f(x)$ attain at the local maxima?
$f\bigl($
$\bigr)=$
,
$f\bigl($
$\bigr)=$
(Enter in same order as above. If rounding, keep at least 3 significant digits.)
Local minima can only occur where $f$ changes from decreasing to increasing (which can happen only at critical points). Find the local minima of $f$.
$f$ has a local minimum at: $x=$
.
What value does $f(x)$ attain at the local minimum?
$f\bigl($
$\bigr)=$
Find the locations and values of the global maximum and minimum of $f$ in the interval $-1 \le x \le 3$. By global maximum and minimum, we mean the largest and smallest value of $f$ in that interval. The global maximum and minimum must occur at the critical points or the endpoints of the interval, so check the values of $f$ at those points to see which is the smallest and largest.
Calculate the value of $f$ at the two critical points that occur in the interval. (Yes, you are repeating yourself.)
$f\bigl($
$\bigr)=$
,
$f\bigl($
$\bigr)=$
(Enter critical points in increasing order. If rounding, keep at least 3 significant digits.)
Calculate the value of $f$ at the two endpoints.
$f(-1)=$
,
$f(3)=$
The global maximum of $f(x)$ in the interval $[-1,3]$ is the largest of those four values.
The global maximum occurs at $x=$
.
The value of the global maximum is $f=$
.
The global minimum of $f(x)$ in the interval $[-1,3]$ is the smallest of those four values.
The global minimum occurs at $x=$
.
The value of the global minimum is $f=$
.
Find the locations and values of the global maximum and minimum of $f$ in the interval $-4 \le x \le 1$.
Calculate the value of $f$ at the two critical points that occur in the interval.
$f\bigl($
$\bigr)=$
,
$f\bigl($
$\bigr)=$
(Enter critical points in increasing order. If rounding, keep at least 3 significant digits.)
Calculate the value of $f$ at the two endpoints.
$f(-4)=$
,
$f(1)=$
The global maximum of $f(x)$ in the interval $[-4,1]$ is the largest of those four values.
The global maximum occurs at $x=$
.
The value of the global maximum is $f=$
.
The global minimum of $f(x)$ in the interval $[-4,1]$ is the smallest of those four values.
The global minimum occurs at $x=$
.
The value of the global minimum is $f=$
.