Question

Help with this mean value theorem question? (attachment)

Answer 1

\[f(x)=x^2\] is a polynomial, therefore continuous and differentiable on the whole real line, so it does satisfy the hypothesis of the mean value theorem

Answer 2

before we finish the second question, we can know what \(c\) is instantly, because \(f(x)=x^2\) is a quadratic, so \(c\) will be right in the middle of the interval \([-6,4]\)
i.e. \(c=-1\)
but we can still find it the way asked for if you like

Answer 3

so the answers -1?

Answer 4

yes
but that is because i happen to know it is right in the center of the interval
should we go ahead and find it? it is not always in the center

Answer 5

ok

Answer 6

we need only a few number so start: \(f(4), f(-6)\) first
since \(f(x)=x^2,f(4)=16,f(-6)=36\)

Answer 7

therefore
\[\frac{f(4)-f(6)}{4-(-6)}=\frac{16-36}{10}=\frac{-20}{10}=-2\]

Answer 8

we also need \(f'(x)=2x\)

Answer 9

now set
\[f'(c)=2c=-2\] and solve for \(c\) to get \(c=-1\) as promised

Answer 10

ok thank you so much!

Answer 11

yw
hope steps were clear