Question

couple questions

1. Find c for Rolle's Theorem for f(x) = x^3 - 12x when 0 is equal to or less than x which is equal to or less than 2 √3

2. Find c for Mean Value Theroem for f(x) = (x - 2) / x on [2, 4]

Answer 1

lets do the second one
first one is the same
you need some numbers
you need \(f(2),f(4)\)

Answer 2

you got those?

Answer 3

yes, i understand doing the f(4) - f(2) / 4 -2. i got that equals 1. got the derivative of (x-2)/x to be 2/x^2.
after i set 0 = 2/c^2-1 i get confused

Answer 4

i don't think the 1 is right

Answer 5

\[f(4)=\frac{4-2}{4}=\frac{2}{4}=\frac{1}{2}\\
f(2)=\frac{2-2}{2}=0\\
f(4)-f(2)=\frac{1}{2}-0=\frac{1}{2}\\
\frac{f(4)-f(2)}{4-2}=\frac{\frac{1}{2}}{2}=\frac{1}{4}\]

Answer 6

oh okay, well i am still confused on what do once i set 0 = 2/c^2 - 1/4?

Answer 7

why the zero? you want
\[\frac{2}{c^2}=\frac{1}{4}\] right?

Answer 8

\[\frac{2}{c^2}=\frac{1}{4}\iff c^2=8\]

Answer 9

that does give you two possible values for \(c\) but only \(2\sqrt2\) is in the interval \([2,4]\)

Answer 10

oh huh, i didnt even think of cross multiplying. was looking too closely at examples my teacher gave i suppose. thank you!
still confused on the first problem though

Answer 11

your function is zero at the endpoints of the interval, which by rolles theorem tells you the derivative must be 0 somewhere in the interval
take the derivative, (which will be a quadratic) set it equal to zero, and solve
if it is not obvious where it is zero, since it is a quadratic, you can use the quadratic formula to solve for it

Answer 12

actually now that i look at it, no quadratic formula necessary
\[f(x)=x^3-12x\\
f'(x)=3x^2-12=3(x^2-4)=3(x+2)(x-2)\] zeros can now easy to find

Answer 13

gotcha. zeros at 2 and -2. Seems like the answer is c =2. thank you for the help again!