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If g is a differentiable function such that g(x)<0 for all real numbers x and if f'(x) = (x2 - 4)g(x), which of the following is true? i. f has relative maxima at x-2 and at x-2.
ii. f has relative minima at x=-2 and at x=2.
iii. It cannot be determined if fhas any relative extrema.
iv. f has a relative maximum at x=-2 and a relative minimum at x=2.
v. f has a relative minimum at x=-2 and a relative maximum at x=2.

Sagot :

facundo3141592

We want to see what can we say about f(x) extremes for the given information. We will see that the correct option is:

V: f has a relative minimum at x=-2 and a relative maximum at x=2.

Maximums and minimums.

A given function f(x) will have a maximum/minimum at x₀ if:

f'(x₀) = 0

  • It will be a maximum if f''(x₀) < 0
  • It will be a minimum if f''(x₀) > 0.

Here we do know that:

  • f'(x) = (x^2 - 4)*g(x)
  • g(x) < 0 for all numbers x.

Then the only zeros of f'(x) are when (x^2 - 4) = 0.

And this happens for x = 2 and x  = -2

Now let's see if these are minimums or maximums, we have:

f''(x) = 2*x*g(x) + (x^2 - 4)*g'(x)

If we evaluate this in x = 2, we get:

f''(2) = 2*2*g(2) + (2^2 - 4)*g'(2)

       = 4*g(2) + 0*g'(2) = 4*g(2)

And g(x) is always negative, then 4*g(2) < 0, then:

f''(2) < 0

Meaning that in x = 2 we have a relative maximum.

For x = -2 we have:

f''(-2) = 2*-2*g(-2) + ((-2)^2 - 4)*g'(-2)

       = -4*g(2) > 0

Then f''(-2) > 0, meaning that in x = -2 we have a relative minimum.

Then the correct option is:

V:  f has a relative minimum at x=-2 and a relative maximum at x=2.

If you want to learn more about maximums and minimums, you can read:

https://brainly.com/question/1938915

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