The maximum value of the function f(x)=e^x+x | vedclass

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Question

(B) Given the function $f(x) = e^x + x \ln x$ on the interval $[1, 2]$.First,we find the derivative of the function:$f'(x) = \frac{d}{dx}(e^x + x \ln

Vedclass · Accepted Answer

$e^2+2 \ln 2$

Vedclass · Answer

(B) Given the function $f(x) = e^x + x \ln x$ on the interval $[1, 2]$.First,we find the derivative of the function:$f'(x) = \frac{d}{dx}(e^x + x \ln x) = e^x + (1 \cdot \ln x + x \cdot \frac{1}{x}) = e^x + \ln x + 1$.For $x \in [1, 2]$,we observe that $e^x > 0$,$\ln x \geq 0$,and $1 > 0$. Thus,$f'(x) > 0$ for all $x \in [1, 2]$.Since the derivative $f'(x)$ is always positive on the interval,the function $f(x)$ is strictly increasing on $[1, 2]$.Therefore,the maximum value occurs at the right endpoint of the interval,which is $x = 2$.Calculating $f(2)$:$f(2) = e^2 + 2 \ln 2$.Thus,the correct option is $B$.

The maximum value of the function $f(x)=e^x+x \ln x$ on the interval $1 \leq x \leq 2$ is

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