True or false: 1) if f'(c)=0, then f has a local maximum or minimum at x=c. i think this one is false, but i'm not sure. 2) if f is continuous on [a, b] and differential on (a, b) and f'(x) = 0 on (a, b), then f is constant on [a, b]. 3) the mean value theorem can be applied to f(x) = 1/x^2 on the interval [-1, 1]. i'm pretty sure the last one is false.

We want to see if each one of the given statements is true or false.

The answers are:

Let's see each statement.

1) "If f'(c)=0, then f has a local maximum or minimum at x=c."

We can find a counterexample, for:

f(x) = x^3

f'(x) = 2*x^2

and

f'(0) = 2*0^2 = 0

but x = 0 is not a maximum nor a minimum in f(x) = x^3, it actually is an inflection point, so the statement is false.

2) "If f is continuous on [a, b] and differential on (a, b) and f'(x) = 0 on (a, b), then f is constant on [a, b]."

We know that if f'(x) = 0, then the function does not depend on the variable x, meaning that the function is constant.

So if we have:

f'(x) = 0 ∀x ∈(a, b)

Then f(x) does not depend on the value of x on the interval (a, b).

Thus the statement is true.

3) "The Mean Value Theorem can be applied to f(x) = 1/x^2 on the interval [-1, 1]."

Yes, this is false, the mean value theorem needs a continuous function, and 1/x^2 has a discontinuity at x = 0, which belongs to the interval [-1, 1].

If you want to learn more, you can read:

link

1. False. does not necessarily mean that is a maximum or minimum. It could just as easily be a saddle point. For example, consider the function with when , yet for and for .

2. True. If for all in , then must be constant on that interval.

3. False. In order for the MVT to apply, must be continuous on the closed interval. But does not exist at .

y/x=18

step-by-step explanation:

5

step-by-step explanation: