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Determine if the function shown below has a global maximum and/or a global minimum.
The function has a global maximum but no global minimum. Since
the function f keeps decreasing, but it never hits bottom.
Find the global max and min of the function f (x) = x2 + 2 on the interval [ -4 , 5 ].
First we find the critical points. The derivative is
f '(x) = 2x
which is zero only when x = 0, so that's our only critical point.
Next we evaluate f at the critical point and at the endpoints of the interval.
The global max of f on [-4, 5] is y = 27, which occurs at x = 5. The global min of f on [-4, 5] is y = 2, which occurs at x = 0.
Remember that there's a difference between the maximum value of a function (a y-value) and where that maximum occurs (an x-value).
On the interval [-3, 3], where is the function f (x) = x3 – 12x largest?
This question is asking us to find the x-value at which the global maximum of f occurs.
First we find the critical points. The derivative
f '(x) = 3x2 – 12
is zero when x = ± 2, so these are our critical points.
Next we evaluate f at the critical points and at the endpoints of the interval.
The maximum value of f occurs at x = 3.
It's possible that we might find a critical point that's not in the interval given. If this happens, ignore the offending critical point. The function f can't possibly have its maximum at x = -2 if we're only looking at the interval 0 < x < 3.
On the interval [0, 3], where is the function f (x) = x3 – 12x smallest?
We found in the example above that f has critical points at x = ± 2. However, x = -2 isn't in the interval we're looking at, so we can toss it out. We only keep the critical point x = 2.
We evaluate the function f at the critical points in [0, 3] and the endpoints of [0,3].
On this interval, the function f is smallest at x = 0.