CPM Homework Help : CALC Problem 6-50

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

$y = 2x e^x$ on $(−∞, ∞)$
Hint 1(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 $y^\prime= 0$ AND where $y^\prime =$ DNE.
Hint 2(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 3(a):
2nd-Derivative Test: Evaluate each candidate in the 2nd-derivative.
If $f^{\prime\prime}$ (candidate) is negative, then you found a local maximum.
If $f^{\prime\prime}$ (candidate) is positive, then you found a local minimum.
If $f^{\prime\prime}$ (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 1(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 2(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.