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

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(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 $[-2, 2]$?

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