Can't find c for Mean Value Theorem

[Kristy]

Verify that the function satisfies the hypotheses of the
Mean Value Theorem on the given interval. Then find all numbers
C that satisfy the conclusion of the MVT.

#12 f(x) = \(\displaystyle x^{3}+x-1\)

[0,2]

My Work for Number 12:

f(x) = \(\displaystyle x^{3}+x-1\) [0,2]
MVT
* f is continuous on [0,2]
* f’(x) = \(\displaystyle 3x^{2}+1\) is the differentiated equation and exists on (0,2)
* There exists a c, 0 < c < 2, such that
f'(b) = \(\displaystyle \frac{F(b)-F(a)}{b-a}\)
Sorry those are supposed to be small f’s not capital F’s.

f(0)= \(\displaystyle 0^{3}+0-1 =-1\)

f(2)= \(\displaystyle 2^{3}+2-1 =8+2-1=9\)

\(\displaystyle 3x^{2}+1 = \frac{F[2]-F[0]}{2-0}\)

\(\displaystyle = \frac{9+1}{2-0} = \frac{10}{2}\)

\(\displaystyle 3x^{2}+1 = 5\)

Now I don’t know what to do next. I think I showed it satisfied the hypotheses of the MVT. I don’t know how to find the c values.

Also why is this ignoring multiple spaces?

 

[galactus]

\(\displaystyle \L\\x^{3}+x-1, \;\ [0,2]\)

f is a polynomial and continuous and differentiable everywhere.

\(\displaystyle \L\\f(a)=f(0)=-1, \;\ f(b)=f(2)=9\)

\(\displaystyle \L\\f'(x)=3x^{2}+1, \;\ f'(c)=3c^{2}+1\)

\(\displaystyle \L\\f'(c)=\frac{f(b)-f(a)}{b-a}\) becomes

\(\displaystyle \L\\3c^{2}+1=5\)

\(\displaystyle \L\\c=\pm\frac{2}{\sqrt{3}}\)

Only \(\displaystyle \L\\\frac{2}{\sqrt{3}}\) is in the interval [0,2].

so, \(\displaystyle c=\frac{2}{\sqrt{3}}\) is the number whose existence is guaranteed by the MVT.

 

[tkhunny]

[0,2]?

 

[Kristy]

[0,2]?

Yes, that is what the problem statement says. Is that unusual or wrong?

Edit: Thanks also for the explanation and graph in the previous post showing why there can only be 1 c value for this time.

 

[tkhunny]

I was just questioning "differentiable everywhere".

 

[Kristy]

I was just questioning "differentiable everywhere".

Polynomials are differentiable everywhere? I just thought so but maybe there's more to it. Why are you questioning it?

 

[tkhunny]

It's Domain is restricted to [0,2]. Are you SURE it's differentiable on [0,2]?

I'm just trying to make you think about it, rather than just write it and move on.

 

[Kristy]

It's Domain is restricted to [0,2]. Are you SURE it's differentiable on [0,2]?

I'm just trying to make you think about it, rather than just write it and move on.

Okay, the only things I could think about are if it had a cusp or sharp point like absolute value function, or if it had a jump discontinuity or an asymptote. I don't t hink a missing point discontinuity would matter.

Did i think about that right? Also for that other problem I did number 7 I just wrote the answer from the back of the book. I don't really undrestand it.