$(1+x)^y=e$
for all real values. Take care to point out all key points and key behaviour.

 Sketch the graph of
$y=\frac{x1}{x+1}$
.  Find the second derivative of
$f(x)=2^{x}.$
 Evaluate
$\lim_{x\to\frac{\pi}{2}} \frac{1}{\tan x}.$
 Sketch the graph of

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:$$ \lim_{x\to\infty}f(x)=0 \\ \lim_{x\to1}f(x)=0 \\ \lim_{x\to0^}f(x)=\infty \\ \lim_{x\to0^+}f(x)=\infty $$
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$\left(1e^{1/2k}, 2k\right).$
We can put all this information together to sketch the graph: