natHenderson
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This is the work I've done so far, I'm allowed to use a graphing calculator but it told me there where two zeroes, which isn't an option:
Mean Value Theorem Problem
- Thread starter natHenderson
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This is the work I've done so far, I'm allowed to use a graphing calculator but it told me there where two zeroes, which isn't an option:
HallsofIvy
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It's not clear to me what you have done! I see that you have written out the polynomial \(\displaystyle \frac{1}{4}x^4-\frac{2}{3}x^3+\frac{1}{2}x^2- \frac{1}{2}x\), then shown that same polynomial evaluated at x= 1.565, divided by 1.565 and set them equal. Why? Do you not know what the "mean value theorem" is? It does NOT say that the polynomial is equal to that quotient. It says that, at some value of x between 0 and 1.565, the derivative of the polynomial is equal to that quotient.
Now, what number did you get for that quotient? The derivative of \(\displaystyle \frac{1}{4}x^4-\frac{2}{3}x^3+\frac{1}{2}x^2- \frac{1}{2}x\) is \(\displaystyle x^3- 2x^2+ x- \frac{1}{2}\). How many values of x, between 0 and 1.565 is that derivative equal to the quotient?
I get -0.39207 for the quotient and when I graph \(\displaystyle y= x^3- 2x^3+ x- \frac{1}{2}\) and \(\displaystyle y= -0.39207\) they cross three times.
Now, what number did you get for that quotient? The derivative of \(\displaystyle \frac{1}{4}x^4-\frac{2}{3}x^3+\frac{1}{2}x^2- \frac{1}{2}x\) is \(\displaystyle x^3- 2x^2+ x- \frac{1}{2}\). How many values of x, between 0 and 1.565 is that derivative equal to the quotient?
I get -0.39207 for the quotient and when I graph \(\displaystyle y= x^3- 2x^3+ x- \frac{1}{2}\) and \(\displaystyle y= -0.39207\) they cross three times.
natHenderson
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Ohh, I forgot it was the derivative, minor mistake, thanks for letting me know!
Steven G
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No, it was not a minor mistake. First you set that quotient to 0 which was not what you needed to do. Then you set the quotient to f(x) which was wrong. You obviously do not know (at least when you were doing this problem) what the MVT states. This was not a minor mistake. Even if you had written the derivative of f(x) instead of f(x) you set the quotient to 0.
MVT states for you problem that if f(x) is continuous on [0, 1.565], then there is at least one c in (0, 1.565) such that {f(1.565) - f(0)}/(1.565-0) = f '(c)
Did you state why f(x) is continuous on [0, 1.565]? No!
Did you use f(0) in the formula? Possibly, since you knew it was 0.
Did you calculate f'(c) and set it equal to the quotient? No
Did you solve for c? No.
These are not minor mistakes, sorry.