For what value of x is the function f(x) = e^ | vedclass

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(A) Given the function $f(x) = e^{ax} + e^{-ax}$.To find where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(e^

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$x > 0$

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(A) Given the function $f(x) = e^{ax} + e^{-ax}$.To find where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(e^{ax} + e^{-ax}) = a e^{ax} - a e^{-ax}$.Factor out $a e^{-ax}$:$f'(x) = a e^{-ax} (e^{2ax} - 1)$.For the function to be increasing,we require $f'(x) > 0$.Since $a > 0$ and $e^{-ax} > 0$ for all real $x$,the sign of $f'(x)$ depends on $(e^{2ax} - 1)$.We set $e^{2ax} - 1 > 0$,which implies $e^{2ax} > 1$.Taking the natural logarithm on both sides:$2ax > \ln(1) = 0$.Since $a > 0$,we divide by $2a$ to get $x > 0$.Thus,the function is increasing for $x > 0$.

For what value of $x$ is the function $f(x) = e^{ax} + e^{-ax}$,where $a > 0$,an increasing function?

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