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If $f(x)=e^{x+1}-1$. Find the sum of all of the values of 'n' that makes $g(x)$ differentiable over $\mathbb{R}$.

$g(x)=100|f(x)|-\sum_{k=1}^n|f(x^k)|$ such that $n \in N$.

I have no idea of how to procced but I can figure out from desmos.com that $f(x)\ge0$ for $x\ge-1$

But could not procced for $\sum_{k=1}^n|f(x^k)|$

MathTrain
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Samar Imam Zaidi
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1 Answers1

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Here is a sequence of hints, in the form of statements to be proven by the reader:

  1. $f(x^k)$ is positive for $k$ even.
  2. $f(x^k)$ crosses zero exactly once for $k$ odd, at $x=-1$.
  3. Hence the derivative of $|f(x^k)|$ has a discontinuity only at $x=-1$ and only for $k$ odd.

For $g$ to be differentiable at $x=-1$, the discontinuity in the derivative of $100 |f(x)|$ must exactly cancel that of the sum.

Lemma: The above holds if and only if $$100 f'(-1) = \sum_{\text{k odd}}^n f(x^k)'|_{x=-1}$$ Proof: Exercise.

Now to answer the question, find a formula for $f(x^k)'|_{x=-1}$ (hint: it's proportional to $k$), evaluate the sum on the right, and find which values of $n$ make it equal to $100 f'(-1)$. Note: There will be two such values of n!

Yly
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