Question

f(x)=(x-3)^-2. Find all values of c in (1,7) such that f(7)-f(1)=f'(c)(7-1)

Answer 1

Okay well we have \(f(x)=(x-3)^{-2}\)
We also know that \(f'(x)=-2(x-3)^{-3}\)
So we can equivocate the statement to:
\[\eqalign{
&f(7)-f(1)=-2(c-3)^{-3}(6) \\
&(7-3)^{-2}-(1-3)^{-2}=-2(c-3)^{-3}(6) \\
&(4)^{-2}-(-2)^{-2}=-12(c-3)^{-3} \\
&\frac{1}{16}-\frac{1}{4}=-\frac{12}{(c-3)^3}
}\]
Now it is just a matter of isolating for c

Answer 2

@KeithAfasCalcLover how do we isolate for c?

Answer 3

Ehh you would have something a little like this:
\[\eqalign{
&\frac{1}{16}-\frac{1}{4}=-\frac{12}{(c-3)^2} \\
&(c-3)^2\left(\frac{1}{16}-\frac{1}{4}\right)=-12 \\
&(c-3)^2=-\frac{12}{\left(\frac{1}{16}-\frac{1}{4}\right)} \\
&c-3=\pm\sqrt{-\frac{12}{\left(\frac{1}{16}-\frac{1}{4}\right)}} \\
&c=3\pm\sqrt{-\frac{12}{\left(\frac{1}{16}-\frac{1}{4}\right)}} \\
}\]