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We have $$f(x) = \frac{3^x}{5^x} + \frac{4^x}{5^x} -1$$ I need to show that it is strictly decreasing, how do I show this?

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3 Answers 3

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Hint: show the derivative $f'$ is negative. ..

So, $f'(x)= \left ( e^{xln {\frac 35}}+ e^{xln{\frac 45}}-1\right ) '=\ln\frac 35 \cdot (\frac 35)^x+\ln \frac 45 \cdot (\frac45)^x\lt 0 , \forall x\in \mathbb R $. ..

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for constant $0 < C < 1,$ $$ \frac{d}{dx} C^x = \frac{d}{dx} e^{x \log C} = C^x \log C < 0 $$

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Note that for $t>0$: $$\left(\frac35\right)^x>\left(\frac35\right)^{x+t} \iff 1>\left(\frac35\right)^t \iff 5^t>3^t.$$

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