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**Vamz** Verify Mean Value Theorem for the function $\displaystyle f(x)=x^2+lnx-1$ on the interval [1,2]

by finding (all) the appropriate point(s) c where the derivative equals the slope of the

secant line between the end points of the interval.

I think I know how to do this. But I have got to solve it without a calculator apparently!!

slope of secant line is $\displaystyle ln2+3$

$\displaystyle f'(x)=2x+\frac{1}{x}$

let m = ln2 + 3

$\displaystyle 2x+\frac{1}{x}=m$

$\displaystyle 2x^2+1=mx$

$\displaystyle 2x^2-mx+1$

$\displaystyle \frac{m\pm\sqrt{(m^2-8)}}{4}$

now substitute:

$\displaystyle \frac{ln2 + 3\pm\sqrt{((ln2 + 3)^2-8)}}{4}$

now I'm stuck here. How do I solve this without a calculator?

Thanks!