Question

find the value(s) of c guaranteed by the mean value theorem for integrals for the function over the given interval.

f(x)=9/x^3 [1,3]

Answer 1

ok that was wrong, let me start again
\[f(3)=\frac{9}{3^3}=\frac{1}{3}\]
\[f(1)=9\]
\[\frac{f(3)-f(1)}{3-1}=\frac{\frac{1}{3}-9}{2}= -\frac{13}{3}\]

Answer 2

then find
\[f'(x)=-\frac{27}{x^4}\] and finally solve
\[-\frac{27}{x^4}=-\frac{13}{3}\] for x

Answer 3

this is what I got F(b)-F(a)=4
9/c^3=2 ....am I right so far?.....how did u get 1/3?

Answer 4

\[f(3)=\frac{9}{3^3}=\frac{1}{3}\] did i make a mistake it is late

Answer 5

oh okay, hold on let me check

Answer 6

you have
\[f(x)=\frac{9}{x^3}\] so i get
\[f(3)=\frac{1}{3}\] and
\[f(1)=9\]

Answer 7

also
\[f'(x)=-\frac{27}{x^4}\]

Answer 8

how did u get that for f'?

Answer 9

in general can you tell me the steps to finding c?

Answer 10

\[f(x)=\frac{9}{x^3}=9x^{-3}\]
\[f'(x)=-27x^{-4}=-\frac{27}{x^4}\]

Answer 11

ok lets go slow
your job is to find
\[\frac{f(b)-f(a)}{b-a}\] which of course is a number
i got
\[-\frac{13}{3}\] as the number

Answer 12

then i found
\[f'(x)=-\frac{27}{x^4}\]

Answer 13

which means
\[f'(c)=-\frac{27}{c^4}\]

Answer 14

last job is to set them equal and solve for "c"

Answer 15

\[-\frac{27}{c^4}=-\frac{13}{3}\] solve for c

Answer 16

\[13c^4=81\]
\[c^4=\frac{81}{13}\]
\[c=\sqrt[4]{\frac{81}{13}}\]

Answer 17

or
\[c=\frac{3}{\sqrt[4]{13}}\]

Answer 18

thanks for the clarification