Question

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.

Answer 1

1. False. $f'(c)=0$ does not necessarily mean that $f(c)$ is a maximum or minimum. It could just as easily be a saddle point. For example, consider the function $f(x)=x^3$ with $f'(x)=3x^2=0$ when $x=0$, yet $f(x)<0$ for $x<0$ and $f(x)>0$ for $x>0$.

2. True. If $f'(x)=0$ for all $x$ in $(a,b)$, then $f$ must be constant on that interval.

3. False. In order for the MVT to apply, $f$ must be continuous on the closed interval. But $\dfrac1{x^2}$ does not exist at $x=0$.

Answer 2

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

The answers are:

• 1) false
• 2) true
• 3) false.

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:

https://brainly.com/question/21447009