Submitted by jetay1 on Mon, 01/23/2017 - 15:28
Forums:
The last part of Q2 asks us to show that $y=3e^{2x}+14\ln\left(\frac{4}{3}-x\right)$ has a minimum between $x=\frac{1}{2}$ and $x=1$. Normally I would do this by differentiating and subbing in the two $x$ values and looking for a change of sign (from negative to positive). But it seems from the way the question is set out that we're supposed to do it by using the Maclaurin series expansion of $e^x$. Do we also need to use the Maclaurin expansion of $\ln (1+x)$? And then maybe differentiate the series expansions, or use some inequality arguments like in the earlier parts?
You are indeed meant to
You are indeed meant to differentiate once (instead of twice) and then examine the signs of the derivative at either side of the extrema. So look at the sign of $y'$ at $x=1/2$ and at $x=1$ using the given inequalities to deduce the sign. Post back with your progress if you're still stuck.
Thanks
It worked out easily. Thank you. Slight pause while figuring out the inequality for $e^2$, but not too much trouble.
No problem!
Well done! It's a neat little question - kudos to you for doing it without too much trouble.