The positive value of $k$ for which the equation $ke^x - x = 0$ has only one root.
The given equation holds true only if $k\in(0,1)$ but I couldn't get anything further than this. The correct answer for this is $k=\frac1e$.
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Is your equation $ke^{x}=0$? – aleden Nov 04 '17 at 15:18
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Are you sure you copied the question correctly? If $k>0$, then $k e^x > 0$ has no roots! – aschepler Nov 04 '17 at 15:18
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sorry! my mistake. i edited it – user496593 Nov 04 '17 at 15:20
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Find the global minimum minimum by solving $$\frac{\partial}{\partial x} (k\exp(x)-x)=0$$ $$k\exp(x)-1=0$$ $$\exp(x)=1/k$$ $$x=\ln(1/k)$$ So the value at the minimum is $$k\exp(\ln(1/k))-\ln(1/k)=1-\ln(1/k)$$ There will be only one root when the global minimum $=0$: $$1-\ln(1/k)=0$$ Solve for k.
Wouter
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also if i have to solve the following equation for 2 distinct roots, i'll have to make the minimum value negative. right? – user496593 Nov 04 '17 at 15:32
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yes. To complete the argument, note that the function is positive in both the $x\rightarrow +\infty$ and the $x\rightarrow +\infty$ limits, so the Intermediate value theorem guarantees the existence of roots. – Wouter Nov 04 '17 at 15:35