If f(x) = begin{cases} -x^3 + 1, & text{if } | vedclass

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

Question

(D) Given the function $f(x) = \begin{cases} -x^3 + 1, & x \leq 1 \ |x - 1| + \lambda, & x > 1 \end{cases}$.At $x = 1$,the value of the function is $

Vedclass · Accepted Answer

$f(x)$ has a point of minima at $x = 1, \forall \lambda > 0$

Vedclass · Answer

(D) Given the function $f(x) = \begin{cases} -x^3 + 1, & x \leq 1 \ |x - 1| + \lambda, & x > 1 \end{cases}$.At $x = 1$,the value of the function is $f(1) = -(1)^3 + 1 = 0$.For $x > 1$,$f(x) = |x - 1| + \lambda = (x - 1) + \lambda$.As $x 	o 1^+$,$f(x) 	o \lambda$.For $x = 1$ to be a point of minima,we must have $f(1) \leq f(x)$ for all $x$ in a neighborhood of $1$.Since $f(1) = 0$,we require $0 \leq \lambda$ for the function to be continuous or to maintain the minimum at $x=1$ relative to the right-hand side limit.Specifically,if $\lambda > 0$,then for $x$ slightly greater than $1$,$f(x) = (x-1) + \lambda > 0 = f(1)$.Thus,$f(x)$ has a point of minima at $x = 1$ for all $\lambda > 0$.

If $f(x) = \begin{cases} -x^3 + 1, & \text{if } -\infty < x \leq 1 \\ |x - 1| + \lambda, & \text{if } x > 1 \end{cases}$,then:

Similar Questions

If $f(x) = \frac{\sin(\pi \cos^2 x)}{3x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $

Consider the function $f(x) = (x - 2) \log_e x$. Then the equation $x \log_e x = 2 - x$

If $f(x) = \begin{cases} \frac{|x - a|}{x - a}, & x \neq a \\ 1, & x = a \end{cases}$,then:

If the function $f(\alpha) = \begin{cases} \frac{1-\cos 6 \alpha}{36 \alpha^2}, & \alpha \neq 0 \\ k, & \alpha=0 \end{cases}$ is continuous at $\alpha=0$,then $k$ is equal to . . . . . . .

Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module