Verify that the function f(x)=x³+x-1 on [0,2]

[lovetolearn]

Verify that the function f(x)=x³+x-1 on [0,2]

Verify that the function f(x)=x³+x-1 on [0,2] satisfies the
hypotheses of the mean value theorem. Then find all the numbers that
satisfy the conclusion of the MVT. Finally explain what your answer
means geometrically.

f'(x)=3x^(2)+1
f'(c)=(f(c)-f(a))/1-a
3c^(2)+1=((f(2)-f(0))/2-0
3c^(2)+1=((2^3)+2-1))-(-1))/2
3c^(2)+1=((8+2-1)+(1))/2
3c^(2)+1=(9+1)/2
3c^(2)+1=10/2
3c^(2)+1=5
3c^(2)-4 ---> c^(2)=4/3 ---> c=plus or minus 2/the square root of 3
c=-2/sqrt3, +2/sqrt3

it is continuous at every point in [0,2] and differentiable at every point in (0,2). There exists two points between 0 and 2 that satisfies the mean value theorem.

 

[tkhunny]

f'(c)=(f(c)-f(a))/1-a

This makes very little sense. What was your intent?
1) You have forgotten your Order of Operations. Add parentheses to the denominator so that it means what you want.
2) f(c)- f(a)?? What is that?

3c^(2)+1=((f(2)-f(0))/2-0

Now you are getting somewhere, but it's still a little odd.
1) You have changed the definition of "c". It used to be an endpoint, now it's some point in the interior.
2) Still with the Order of Operations. Luckily, that zero on the end saved you.

3c^(2)+1=((2^3)+2-1))-(-1))/2

You really have to be more careful. Count those parentheses.

3c^(2)+1=((8+2-1)+(1))/2
3c^(2)+1=(9+1)/2
3c^(2)+1=10/2
3c^(2)+1=5

Okay, I follow this.

3c^(2)-4 ---> c^(2)=4/3

Two problems again:
1) Why did you bother to move everything to the left-hand side if you were just going to move some of it back to the right-hand side?
2) You had an equation. Where did it go? Oh, now it's back, after the arrow.

---> c=plus or minus 2/the square root of 3
c=-2/sqrt3, +2/sqrt3

it is continuous at every point in [0,2] and differentiable at every point in (0,2). There exists two points between 0 and 2 that satisfies the mean value theorem.

Not quite. Do you really believe -2/sqrt(3) is in (0,2)?

Summary: You've got good stuff going on. You do seem to to have the idea. There's just a whole lot of distraction going on.

Do another one and let's see it a little cleaner.

 

[HallsofIvy]

Verify that the function f(x)=x³+x-1 on [0,2] satisfies the
hypotheses of the mean value theorem. Then find all the numbers that
satisfy the conclusion of the MVT. Finally explain what your answer
means geometrically.

f'(x)=3x^(2)+1
f'(c)=(f(c)-f(a))/1-a
3c^(2)+1=((f(2)-f(0))/2-0
3c^(2)+1=((2^3)+2-1))-(-1))/2

You never do "verify that the function ... satisfies the hypotheses of the mean value theorem". What are those hypotheses?

3c^(2)+1=((8+2-1)+(1))/2
3c^(2)+1=(9+1)/2
3c^(2)+1=10/2
3c^(2)+1=5
3c^(2)-4 ---> c^(2)=4/3 ---> c=plus or minus 2/the square root of 3
c=-2/sqrt3, +2/sqrt3

it is continuous at every point in [0,2] and differentiable at every point in (0,2). There exists two points between 0 and 2 that satisfies the mean value theorem.