The function f(x) = x e^{x(1-x)}, x in R,is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given function is $f(x) = x e^{x-x^2}$.To find the intervals of increase and decrease,we calculate the derivative $f'(x)$:$f'(x) = 1 \cdot e^{x-x^

Vedclass · Accepted Answer

increasing in $\left(-\frac{1}{2}, 1\right)$

Vedclass · Answer

(A) Given function is $f(x) = x e^{x-x^2}$.To find the intervals of increase and decrease,we calculate the derivative $f'(x)$:$f'(x) = 1 \cdot e^{x-x^2} + x \cdot e^{x-x^2} \cdot (1-2x)$$f'(x) = e^{x-x^2} [1 + x - 2x^2]$$f'(x) = e^{x-x^2} [-(2x^2 - x - 1)]$$f'(x) = -e^{x-x^2} (2x+1)(x-1)$.For the function to be increasing,$f'(x) > 0$:$-e^{x-x^2} (2x+1)(x-1) > 0$$(2x+1)(x-1) The roots are $x = -\frac{1}{2}$ and $x = 1$. The inequality holds for $x \in \left(-\frac{1}{2}, 1\right)$.Thus,the function is increasing in $\left(-\frac{1}{2}, 1\right)$.

The function $f(x) = x e^{x(1-x)}, x \in R$,is

Similar Questions

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases is

For what values of $x$ is the function $f(x) = x - \log x$ strictly decreasing?

The set of all real values of $a$ such that the real valued function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$ is strictly increasing in its entire domain is

In which interval is the given function $f(x) = 2x^3 - 15x^2 + 36x + 1$ monotonically decreasing?

Let $\lambda^{}$ be the largest value of $\lambda$ for which the function $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$ is increasing for all $x \in \mathbb{R}$. Then $f_{\lambda^{}}(1) + f_{\lambda^{*}}(-1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module