0 votes 0 votes The value of the real variable $\mathrm{x} \geq 0$, which maximizes the function $f(x)=$ $x^{e} e^{-x}$ is ___________ (up to two decimal places) Others gateda-sample-paper-2024 + – admin asked Oct 21, 2023 • edited Oct 25, 2023 by makhdoom ghaya admin 1.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes an easier way to find derivative $y = ln(f(x)) = e \cdot ln(x) \ – \ x $ $ y’ = \frac{f’(x)}{f(x)} = e/x \ – \ 1 $ $f’(x) = f(x) \cdot (e/x – 1) = 0$ $x = e$ ssingla answered Dec 13, 2023 ssingla comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes $f'(x)=f(x)(1/x-1/e)=0$ x=e sutirtha answered Dec 10, 2023 sutirtha comment Share Follow See all 4 Comments See all 4 4 Comments reply 6rivu commented Dec 12, 2023 reply Follow Share Is it taylor’s expansion? If not, could you please please refer the topic in which this concept comes in? 0 votes 0 votes sutirtha commented Dec 12, 2023 reply Follow Share It is simply obtained by taking the first derivative of the given function by using the product (chain) rule. 1 votes 1 votes 6rivu commented Dec 12, 2023 reply Follow Share Please upload a snapshot of the full solution if that’s possible. Thanks! 0 votes 0 votes rajveer43 commented Jan 5 reply Follow Share chech out this solution https://gateoverflow.in/413573/gate-science-artificial-intelligence-sample-paper-question?show=418149#a418149 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Let \(f(x) = x^e \cdot e^{-x}\). Find the derivative \(f'(x)\): \[f'(x) = e \cdot x^{e-1} \cdot e^{-x} - x^e \cdot e^{-x}\] \[f'(x) = e^{-x} \cdot x^{e-1} \cdot (e - x)\] Set \(f'(x) = 0\) to find critical points: \[e^{-x} \cdot x^{e-1} \cdot (e - x) = 0\] This equation holds true when \(x = 0\) or \(x = e\). Check the second derivative to determine local maxima: \[f''(x) = e^{-x} \cdot x^{e-2} \cdot (e^2 - 2ex + x^2)\] \[f''(0) = 0 \quad (\text{inconclusive})\] \[f''(e) = -e^{-e} \cdot e^{e-2} = -\frac{1}{e} < 0\] This indicates a local maximum at \(x = e\). Consider boundary points: \[f(0) = 0^e \cdot e^{-0} = 0\] \[f(e) = e^e \cdot e^{-e} = 1\] Compare function values: \[f(0) = 0 < f(e) = 1\] Therefore, the value of \(x\) that maximizes \(f(x) = x^e \cdot e^{-x}\), within the constraint \(x \geq 0\), is \(x = e\), which is approximately $2.72$ (up to two decimal places). So the answer is $2.72$ rajveer43 answered Jan 5 rajveer43 comment Share Follow See all 0 reply Please log in or register to add a comment.