# 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.

## Related topics

## Warm-up Questions

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

## Hints

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