The solution set of f'(x) g'(x),where f(x) | vedclass

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(D) Given $f(x) = \frac{1}{2} \cdot 5^{2x+1} = \frac{5}{2} \cdot 5^{2x}$.Calculating the derivative: $f'(x) = \frac{5}{2} \cdot 5^{2x} \cdot \ln 5 \cd

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$x > 0$

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(D) Given $f(x) = \frac{1}{2} \cdot 5^{2x+1} = \frac{5}{2} \cdot 5^{2x}$.Calculating the derivative: $f'(x) = \frac{5}{2} \cdot 5^{2x} \cdot \ln 5 \cdot 2 = 5 \cdot 5^{2x} \ln 5$.Given $g(x) = 5^x + 4x \ln 5$.Calculating the derivative: $g'(x) = 5^x \ln 5 + 4 \ln 5 = \ln 5 (5^x + 4)$.We need to solve $f'(x) > g'(x)$:$5 \cdot 5^{2x} \ln 5 > \ln 5 (5^x + 4)$.Since $\ln 5 > 0$,we can divide by $\ln 5$:$5 \cdot (5^x)^2 > 5^x + 4$.Let $t = 5^x$. Since $x \in \mathbb{R}$,$t > 0$.The inequality becomes $5t^2 - t - 4 > 0$.Factoring the quadratic: $(5t + 4)(t - 1) > 0$.Since $t > 0$,$5t + 4$ is always positive.Thus,we must have $t - 1 > 0$,which implies $t > 1$.Substituting back $t = 5^x$: $5^x > 1$.Since $5^x$ is an increasing function,$x > 0$.

The solution set of $f'(x) > g'(x)$,where $f(x) = \frac{1}{2} (5^{2x+1})$ and $g(x) = 5^x + 4x \ln 5$ is:

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