If g is differentiable function such that g(x)<0 for all real numbers x and f ' (x)= (x² - 4) g(x), which of the following is true?

f is clearly differentiable everywhere, since we are given its derivative, which is defined everywhere. Not only that, f is twice differentiable since its derivative is the product of two differentiable functions. Let’s calculate it now:

f”(x) = (x² – 4)g'(x) + 2xg(x)

Now, let’s figure out the question. How do we test for relative extrema? Well, we have two methods: the first and second derivative test. The second derivative test is, generally, easier, so we’ll do that first.

First thing to do is find where f'(x) = 0. We have:

(x² – 4)g(x) = 0

Since g(x) < 0 for all x, we may divide both sides by g(x): x² - 4 = 0 x = +2 or -2 These are the only possible zeroes for f'(x). If we test them, we see immediately that they check out. Now, we can determine the nature of these stationary points, by substituting into the second derivative: f''(x) = (x² - 4)g'(x) + 2xg(x) f''(2) = (2² - 4)g'(2) + 2(2)g(2) = 4g(2) f''(-2) = ((-2)² - 4)g'(-2) + 2(-2)g(-2) = -4g(-2) Now, g(x) < 0 for all x, so: f''(2) = 4g(2) < 0 f''(2) = -4g(-2) > 0

Hence, by second derivative test, we have exactly two relative extrema: a maximum at x = 2 and a minimum at x = -2.

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However, it turns out, interestingly enough, that the condition that g is differentiable is superfluous. We don’t need to use the second derivative test. We can use the first derivative test, which requires no differentiation of g(x).

We already know the only two possible relative extrema: x = 2 or -2. Let’s examine x = 2. Suppose x > 2. Then:

x > 2

x² > 4

x² – 4 > 0

So, if x > 2, then f'(x) is a product of a positive function (x² – 4) and a negative function (g(x)). Thus f'(x) < 0. Suppose x < 2, but x > 0. Then:

0 < x < 2 x² < 4 x² - 4 < 0 So f'(x) is the product of two negative functions, so f'(x) > 0. This means that, by the first derivative test, f has a relative maximum at x = 2. Similarly, we can address x = -2:

x < -2
x² > 4

x² – 4 > 0

f'(x) < 0
-2 < x < 0
x² < 4
x² - 4 < 0
f'(x) > 0

Therefore, f(x) has a relative minimum at x = 2.

See? No differentiable g is necessary!

g(6)>24 – that is all we can say. lou h, I think you are wrong because g(5)<=[g(6)+g(4)]/2, so g(6)>24, not 48. It can be obviously seen if we draw a line with point (4; 12) and (5; 18) and then just a bit “pull” it down, so that g” becomes above 0 (for the line it is zero). Then we can get any g(6) which is greater than 24.

You won’t learn it, if you put your homework questions online. That’s not the point of doing homework.

Read your book, I bet you have one.

B

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