If f:[-3,2] rightarrow [0, sqrt[3]{x}] is an | vedclass

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

Question

(C) Given the function $f:[-3,2] ightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \ n^{2/3}, & -1 <

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given the function $f:[-3,2] ightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \ n^{2/3}, & -1 For the function to be onto, the range of $f(n)$ must be equal to the codomain $[0, \sqrt[3]{x}]$.First, evaluate the function at the boundaries and critical points:For $-3 \leq n \leq -1$, $f(n)$ increases from $f(-3) = 2 + \sqrt[3]{-3} \approx 0.55$ to $f(-1) = 2 + (-1)^{1/3} = 2 - 1 = 1$.For $-1 Since the function is onto, the maximum value of the function must be the upper bound of the codomain.The maximum value is $\max(f(-1), f(2)) = \max(1, \sqrt[3]{4}) = \sqrt[3]{4}$.Thus, $\sqrt[3]{x} = \sqrt[3]{4}$, which implies $x = 4$.

If $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ is an onto function defined by $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$, then $x=$

Similar Questions

If $f$ is a real-valued function from $A$ onto $B$ defined by $f(x) = \frac{1}{\sqrt{|x - |x||}}$, then $A \cap B = $

The range of the real valued function $f(x) = \frac{x^2 + 2x - 15}{2x^2 + 13x + 15}$ is

If $f: R \rightarrow R$ is defined by $f(x) = [\frac{x}{5}]$ for $x \in R$, where $[y]$ denotes the greatest integer not exceeding $y$, then $\{f(x) : |x| < 71\}$ is equal to

If $f: R-\{2\} \rightarrow R$ is a function defined by $f(x)=\frac{x^2-4}{x-2}$,then the range is

Domain of the function $f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module