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# Practice Paper 4 Question 13

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

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## Warm-up Questions

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

## Hints

• Hint 1
Have you tried rearranging the equation into a form that is easier to work with?
• Hint 2
Try exploring what happens to $$y$$ near some interesting values of $$x.$$
• Hint 3
Are there any other key points on this graph?
• Hint 4
... in particular, any stationary or inflection points?
• Hint 5
Remember the original equation is given in a different form, and you may have made an assumption about $$x$$ when simplifying it.
• Hint 6
Does this permit any more values of $$x$$ and $$y?$$
• Hint 7
... more specifically, when $$x<-1.$$

## Solution

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.

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 $$1+x=\pm\sqrt[2k]{e}.$$ We choose the negative solution, as that is what's still unaccounted for by the original method, so we get $$\left(-1-\sqrt[2k]{e}, 2k\right).$$

We can put all this information together into a sketch resembling the graph:

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