Consider the function f(x) = frac{x^3}{4} - s | vedclass

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

Question

(B, C) The function $f(x) = \frac{x^3}{4} - \sin(\pi x) + 3$ is continuous on the interval $[-2, 2]$.Evaluating at the endpoints:$f(-2) = \frac{(-2)^3

Vedclass · Accepted Answer

$f(x)$ takes on the value $3 \frac{1}{4}$ in the interval $[-2, 2]$.

Vedclass · Answer

(B, C) The function $f(x) = \frac{x^3}{4} - \sin(\pi x) + 3$ is continuous on the interval $[-2, 2]$.Evaluating at the endpoints:$f(-2) = \frac{(-2)^3}{4} - \sin(-2\pi) + 3 = \frac{-8}{4} - 0 + 3 = -2 + 3 = 1$.$f(2) = \frac{2^3}{4} - \sin(2\pi) + 3 = \frac{8}{4} - 0 + 3 = 2 + 3 = 5$.By the Intermediate Value Theorem, since $f(x)$ is continuous on $[-2, 2]$, it must attain every value in the interval $[f(-2), f(2)]$, which is $[1, 5]$.Since $2 \frac{1}{3} = \frac{7}{3} \approx 2.33$ and $3 \frac{1}{4} = 3.25$ both lie within the interval $[1, 5]$, the function $f(x)$ attains both these values.Thus, both options $B$ and $C$ are correct.

Consider the function $f(x) = \frac{x^3}{4} - \sin(\pi x) + 3$. Which of the following statements is true regarding the values attained by $f(x)$ in the interval $[-2, 2]$?

Similar Questions

If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $

Let $f(x) = x^3$, $x \in [-1, 1]$. Then which of the following are correct?

If a real valued function $f(x) = \begin{cases} e^{\frac{\sin a(x-[x])}{x-[x]}}, & \text{if } x < 1 \\ b+1, & \text{if } x = 1 \\ \frac{|x^2+x-2|}{x-1}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$,then $b \sin a =$ ([x] denotes the greatest integer function)

If function $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$,then $f(x)$ is continuous at how many points?

If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module