Rolle's Theorem and solving for f'(x)=0

[Kristy]

Could you help me to figure out how to solve for the values of x where f ’(x) = 0 for
Number 2.

Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.

\(\displaystyle f(x) = x^{3} - 3x^{2} + 2x + 5\) for [0,2]

My work for number 2:
\(\displaystyle f(x) = x^{3} - 3x^{2} + 2x + 5\) for [0,2]
1) f(x) is continuous on [0,2] as it is a polynomial.
2) f is differentiable on (0,2)
\(\displaystyle f'(x)= 3x^{2} -6x+2\)
3)f(0) = f(2)

\(\displaystyle f(0) = 0^{3} -3(0)^{2} + 2(0) + 5\)
f{0) = 0-0+0+5
f(0) = 5

\(\displaystyle f(2) = 2^{3} -3(2)^{2} + 2(2) + 5\)
f(2)=8-12+4+5
f(2)=5


*The function satisfies the three hypotheses of Rolle’s Theorem.
f'(x)=0
\(\displaystyle f’(x)=3x^{2}-6x+2\) \(\displaystyle \hspace{100}\)for 0<x<2

\(\displaystyle 3x^{2}-6x+2 = 0\)

\(\displaystyle x= \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}\)

\(\displaystyle x= \frac{6 \pm \sqrt{6^{2} -4(2)(3)}}{2(2)}\)

\(\displaystyle x= \frac{6 \pm \sqrt{12}}{4}\)

\(\displaystyle x= \frac{6 \pm \sqrt{(4)(3)}}{4}\)

\(\displaystyle x= \frac{6 \pm 2\sqrt{(3)}}{4}\)

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

\(\displaystyle x= \frac{3 + \sqrt{(3)}}{2}\) = 2.36603
or
\(\displaystyle x= \frac{3 - \sqrt{(3)}}{2}\) = 0.633975

I think the only answer is x = c = \(\displaystyle \frac{3 - \sqrt{(3)}}{2}\) = 0.633975

Does this seem right at all?

 

[morson]

Yes (to me anyway). Nice work.

 

[Deleted member 4993]

Could you help me to figure out how to solve for the values of x where f ’(x) = 0 for
Number 2.

Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.

\(\displaystyle f(x) = x^{3} - 3x^{2} + 2x + 5\) for [0,2]

*The function satisfies the three hypotheses of Rolle’s Theorem.
f'(x)=0
\(\displaystyle f’(x)=3x^{2}-6x+2\) \(\displaystyle \hspace{100}\)for 0<x<2

\(\displaystyle 3x^{2}-6x+2 = 0\)

\(\displaystyle x= \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}\)

\(\displaystyle x= \frac{6 \pm \sqrt{6^{2} -4(2)(3)}}{2(2)}\) ...This should be 2*3 in the denominator
\(\displaystyle x= \frac{6 \pm \sqrt{12}}{4}\)

\(\displaystyle x= \frac{6 \pm \sqrt{(4)(3)}}{4}\)

\(\displaystyle x= \frac{6 \pm 2\sqrt{(3)}}{4}\)

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

\(\displaystyle x= \frac{3 + \sqrt{(3)}}{2}\) = 2.36603
or
\(\displaystyle x= \frac{3 - \sqrt{(3)}}{2}\) = 0.633975

I think the only answer is x = c = \(\displaystyle \frac{3 - \sqrt{(3)}}{2}\) = 0.633975

Does this seem right at all?

 

[Kristy]

Could you help me to figure out how to solve for the values of x where f ’(x) = 0 for
Number 2.

Verify that the function satisfies the three hypotheses of Rolle’s Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.

\(\displaystyle f(x) = x^{3} - 3x^{2} + 2x + 5\) for [0,2]

My work for number 2:
\(\displaystyle f(x) = x^{3} - 3x^{2} + 2x + 5\) for [0,2]
1) f(x) is continuous on [0,2] as it is a polynomial.
2) f is differentiable on (0,2)
\(\displaystyle f'(x)= 3x^{2} -6x+2\)
3)f(0) = f(2)

\(\displaystyle f(0) = 0^{3} -3(0)^{2} + 2(0) + 5\)
f{0) = 0-0+0+5
f(0) = 5

\(\displaystyle f(2) = 2^{3} -3(2)^{2} + 2(2) + 5\)
f(2)=8-12+4+5
f(2)=5


*The function satisfies the three hypotheses of Rolle’s Theorem.
f'(x)=0
\(\displaystyle f’(x)=3x^{2}-6x+2\) \(\displaystyle \hspace{100}\)for 0<x<2

\(\displaystyle 3x^{2}-6x+2 = 0\)

\(\displaystyle x= \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}\)

\(\displaystyle x= \frac{6 \pm \sqrt{6^{2} -4(2)(3)}}{2(3)}\)

\(\displaystyle x= \frac{6 \pm \sqrt{12}}{6}\)

\(\displaystyle x= \frac{6 \pm \sqrt{(4)(3)}}{6}\)

\(\displaystyle x= \frac{6 \pm 2\sqrt{(3)}}{6}\)

\(\displaystyle x= \frac{3 \pm \sqrt{(3)}}{3}\)

\(\displaystyle x= \frac{3 + \sqrt{(3)}}{3}\) = 1.57735
or
\(\displaystyle x= \frac{3 - \sqrt{(3)}}{3}\) = 0.42265

So now there are two c values? c = 1.57735 and 0.42265?
Doess that seem like I fixed it more right?

 

[galactus]

I don't understand your quandary. You found the two values which satisfy the MVT and you found f'(x)=0.

 

[Kristy]

I don't understand your quandary. You found the two values which satisfy the MVT and you found f'(x)=0.

I guess I just wasn't sure what I have to do next. In our notes it stops there.


Edit: Wait a minute! Maybe I did that problem and did all the things I needed to? I didn't even know it but I looked at it again and it seems like I did it? :shock:

 

[galactus]

Yes, it would appear.

 

[Kristy]

Yes, it would appear.

Wow. I guess I just assumed I was wrong.
Also is it bettre to start a new topic about my other question about that graph problem since it seems to have become more of a topic of how to make images show up than answering my actual quewstion on the problem itself? It jsut seems like that issue made it so that the focus.