CPM Homework Help : APCALC Problem 6-50

Write the first and second derivatives of each function. Use these derivatives to test for maxima and minima in the given interval. Remember to check the endpoints.

$y = 2 x e ^ { x }$ on $( - \infty , \infty )$

Hint (a):

Since this is an OPEN interval, you are checking for LOCAL maxima and minima only.

Steps (a):

You can use the 1st- or the 2nd-Derivative Test.
No matter which test you choose, your first step will be to find extrema CANDIDATES by determining where $y' = 0$ AND where $y' =$ DNE.

Hint (a):

1st-Derivative Test: Check values to the left and the right of each candidate.
If $f ^\prime(x)$ changes from positive to negative, then you found a local maximum.
If $f ^\prime(x)$ changes from negative to positive, then you found a local minimum.
If $f ^\prime(x)$ does not change sign, then you found a point of inflection.

Hint (a):

2nd-Derivative Test: Evaluate each candidate in the 2nd-derivative.
If $f ^{\prime\prime}(\text{candidate})$ is negative, then you found a local maximum.
If $f ^{\prime\prime}(\text{candidate})$ is positive, then you found a local minimum.
If $f ^{\prime\prime}(\text{candidate})=0$ or DNE, then this test is inconclusive.

Answer (a):

There is a local minimum at $x = −1$ .
But recall that a local min or max is a $y$ -value, not an $x$ -value.

$y = \frac { x } { x ^ { 2 } + 1 }$ on $[ - 2,2 ]$

Hint (b):

Since this is a CLOSED interval, you are checking for both LOCAL and GLOBAL maxima and minima.

Step 1(b):

Find the local maxima and minima. (Refer to the hints in part (a) for guidance.)
Note: These are also global candidates.

Step 2(b):

Find the global max and min by evaluating the y-value of each candidate.
The highest $y$ -value is the global maximum.
The lowest $y$ -value is the global minimum.

Hint (b):

It is very important to consider candidates where
$y^\prime =$ DNE.
Recall that a derivative will not exist at a cusp, endpoint, jump, hole or vertical tangent.
SO CONSIDER THE ENDPOINTS AS CANDIDATES.