1. ## Mean Value Theorem

Verify Mean Value Theorem for the function $f(x)=x^2+lnx-1$ 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 $ln2+3$
$f'(x)=2x+\frac{1}{x}$
let m = ln2 + 3
$2x+\frac{1}{x}=m$
$2x^2+1=mx$
$2x^2-mx+1$
$\frac{m\pm\sqrt{(m^2-8)}}{4}$
now substitute:
$\frac{ln2 + 3\pm\sqrt{((ln2 + 3)^2-8)}}{4}$

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

Thanks!

2. Originally Posted by Vamz
Verify Mean Value Theorem for the function $f(x)=x^2+lnx-1$ 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 $ln2+3$
$f'(x)=2x+\frac{1}{x}$
let m = ln2 + 3
$2x+\frac{1}{x}=m$
$2x^2+1=mx$
$2x^2-mx+1$
$\frac{m\pm\sqrt{(m^2-8)}}{4}$
now substitute:
$\frac{ln2 + 3\pm\sqrt{((ln2 + 3)^2-8)}}{4}$

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

Thanks!
the value of $x$ in the interval $[1,2]$ that satisfies the conclusion of the MVT is $x = \frac{\ln{2}+3 + \sqrt{(\ln{2}+3)^2 - 8}}{4}$ because $\frac{\ln{2}+3}{4} < 1$