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Let f (x) = xex. If possible, use the second derivative test to determine if each critical point is a maximum, a minimum, or neither.
he first derivative is
f '(x) = xex + ex = ex(x + 1),
so there is one critical point at x = -1. For the second derivative test we need the second derivative, which we can find using the product rule.
At the critical point, we have
Therefore f (x) is concave up at x = -1, so it looks like this:
We must have a minimum at x = -1.
Let f (x) = -x3 + 3x2 – 3x. If possible, use the second derivative test to determine if each critical point is a minimum, maximum, or neither.
The derivative is
This is only zero when x = 1, and never undefined, so x = 1 is the only critical point. The second derivative, using the chain rule, is
f "(x) = -6(x – 1)(1) = -6(x – 1).
f "(1) = 0,
we can't use the second derivative test to determine what type of critical point there is at x = 1. We could use the first derivative test though.
Let f (x) = 5 – x2. If possible, use the second derivative test to determine if each critical point is a minimum, maximum, or neither.
The first derivative is
f '(x) = -2x,
which is 0 when x = 0 so we have a critical point at x = 0. The second derivative is
f "(x) = -2.
Since f "(0) = -2 < 0,
the function f is concave down and we have a maximum at x = 0. The graph confirms this:
When doing these problems, remember that we don't need to know the value of the second derivative at each critical point: we only need to know the sign of the second derivative. This might save us some calculation time.