Sketch $(1+x)^y=e$ for all real values. Take care to point out all key points and key behaviour.
  1. Sketch the graph of $y=\frac{x-1}{x+1}$.
  2. Find the second derivative of $f(x)=2^{x}.$
  3. Evaluate $\lim_{x\to\frac{\pi}{2}} \frac{1}{\tan x}.$

Have you tried rearranging the equation into a form that is easier to work with?

Try exploring what happens to $y$ near some interesting values of $x.$

Are there any other key points on this graph?

… in particular, any stationary or inflection points?

Remember the original equation is given in a different form, and you may have made an assumption about $x$ when simplifying it.

Does this permit any more values of $x$ and $y?$

… more specifically, when $x<-1.$

We first rearrange the equation into something we can work with more easily. Assuming $x+1>0$ and taking log on both sides gives us $\ln{(x+1)^y}=1.$ We can write $\ln{(x+1)^y}$ as $y\ln{(x+1)}$ so the original equation can be expressed as $f(x)=\frac{1}{\ln{(x+1)}}.$

Now let’s see what happens around interesting values of $x,$ namely as it tends to $\infty,$ $-1$ and $0.$ We can find all these limits by seeing what happens to $\ln(x+1)$ at these values (a sketch may help). The results are: $$ \lim_{x\to\infty}f(x)=0 \\ \lim_{x\to-1}f(x)=0 \\ \lim_{x\to0^-}f(x)=-\infty \\ \lim_{x\to0^+}f(x)=\infty $$ Next we take derivatives to identify any stationary and inflection points. The first derivative is $f'(x)=\frac{1}{(x+1)\ln^2(x+1)},$ which has no zeroes, hence no finite minima or maxima.

The second derivative is $f''(x)= \frac{2\ln(x+1)+\ln^2(x+1)}{(x+1)^2\ln^4(x+1)} = \frac{\ln(x+1)+2}{(x+1)^2\ln^3(x+1)}.$ This has a zero at $x=e^{-2}-1.$ Substituting this back into the original equation, we get $y=-1/2,$ so we know the graph has an inflection point at $\left(e^{-2}-1,-1/2\right).$

For $x+1 \lt 0,$ $(x+1)^y > 0$ if and only if $y$ is even. Let $y = 2k.$ We take the $(2k)^{th}$ root of $e$ in the initial equation to get $1+x=\pm e^{1/2k}.$ We choose the negative solution, as that is what’s still unaccounted for by the original method, so we get $\left(-1-e^{1/2k}, 2k\right).$

We can put all this information together to sketch the graph: