GATE Data Science and Artificial Intelligence 2024 | Sample Paper | Question: 54

Question

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)

Answer 1

an easier way to find derivative

$y = ln(f(x)) =  e \cdot ln(x) \ –  \ x $

$ y’ = \frac{f’(x)}{f(x)} = e/x \ – \ 1 $

$f’(x) = f(x)  \cdot (e/x – 1) = 0$

$x = e$

Answer 2

$f'(x)=f(x)(1/x-1/e)=0$

x=e

Comments

Is it taylor’s expansion? If not, could you please please refer the topic in which this concept comes in?

---

It is simply obtained by taking the first derivative of the given function by using the product (chain) rule.

---

Please upload a snapshot of the full solution if that’s possible. Thanks!

---

chech out this solution https://gateoverflow.in/413573/gate-science-artificial-intelligence-sample-paper-question?show=418149#a418149

Answer 3

1. Let \(f(x) = x^e \cdot e^{-x}\). Find the derivative \(f'(x)\): \[f'(x) = e \cdot x^{e-1} \cdot e^{-x} - x^e \cdot e^{-x}\] \[f'(x) = e^{-x} \cdot x^{e-1} \cdot (e - x)\]

2. 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 \(x = e\).

3. 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 \(x = e\).

4. Consider boundary points: \[f(0) = 0^e \cdot e^{-0} = 0\] \[f(e) = e^e \cdot e^{-e} = 1\]

5. 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 $2.72$