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Let $f(x)=x2^x$. Find all values of $x$ such that $f(x)=2f'(x)$.

Aug 10, 2022

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Hi Guest!

$$f(x)=x2^x \\ \implies f'(x)=\dfrac{d}{dx}[x2^x]=\dfrac{d}{dx}(x)*2^x+x\dfrac{d}{dx}2^x \hspace{0.5cm} \text{(Product rule)} \\ \implies f'(x)=2^x+x2^x ln(2)$$

Next, we set: $$f(x)=2f'(x)$$ to get:

$$x2^x=2[x2^xln(2)+2^x] \\ \iff x=2ln(2)x+2 \\ \iff x-ln(4)x=2 \\ x=\dfrac{2}{1-ln(4)}$$

which is the only real solution of x.

I hope this helps.

Aug 11, 2022