### Home > CALC > Chapter 6 > Lesson 6.1.4 > 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. Homework Help ✎

*y =*2*xe*on (−∞, ∞)^{x}*y*=on [ −2, 2]

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

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.

1st-Derivative Test: Check values to the left and the right of each candidate.

If *f* '(*x*) changes from positive to negative, then you found a local maximum.

If *f* '(*x*) changes from negative to positive, then you found a local minimum.

If *f* '(*x*) does not change sign, then you found a point of inflection.

2nd-Derivative Test: Evaluate each candidate in the 2nd-derivative.

If *f* ''(candidate) is negative, then you found a local maximum.

If *f* ''(candidate) is positive, then you found a local minimum.

If *f* ''(candidate) = 0 or DNE, then this test is inconclusive.

There is a local minimum at *x* = −1.

But recall that a local min or max is a *y*-value, not an *x*-value.

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

Find the local maxima and minima. (Refer to the hints in part (a) for guidance.)

Note: These are also global candidates.

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.

It is very important to consider candidates where *y*' = DNE.

Recall that a derivative will not exist at a cusp, endpoint, jump, hole or vertical tangent.

SO CONSIDER THE ENDPOINTS AS CANDIDATES.