If f:[-6,6] rightarrow R is defined by f(x)=x | vedclass

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

Question

(A) Given, $f(x)=x^2-3$. First, we calculate $(f \circ f \circ f)(-1)$: $f(-1) = (-1)^2 - 3 = -2$ $f(f(-1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(-1))) = f

Vedclass · Accepted Answer

$f(4 \sqrt{2})$

Vedclass · Answer

(A) Given, $f(x)=x^2-3$. First, we calculate $(f \circ f \circ f)(-1)$: $f(-1) = (-1)^2 - 3 = -2$ $f(f(-1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(-1))) = f(1) = (1)^2 - 3 = -2$. Next, we calculate $(f \circ f \circ f)(0)$: $f(0) = (0)^2 - 3 = -3$ $f(f(0)) = f(-3) = (-3)^2 - 3 = 6$ $f(f(f(0))) = f(6) = (6)^2 - 3 = 33$. Next, we calculate $(f \circ f \circ f)(1)$: $f(1) = (1)^2 - 3 = -2$ $f(f(1)) = f(-2) = (-2)^2 - 3 = 1$ $f(f(f(1))) = f(1) = (1)^2 - 3 = -2$. Summing these values: $(f \circ f \circ f)(-1) + (f \circ f \circ f)(0) + (f \circ f \circ f)(1) = -2 + 33 - 2 = 29$. Now, checking the options: $f(4 \sqrt{2}) = (4 \sqrt{2})^2 - 3 = 32 - 3 = 29$. Thus, the expression is equal to $f(4 \sqrt{2})$.

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

Similar Questions

$f(x) = (20 - x^4)^{1/4}$ for $0 < x < \sqrt{5}$, then $f(f(1/2))$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$, then the values of $x$ such that $g(f(x))=8$ are

Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are onto,then $g \circ f: A \rightarrow C$ is also onto.

Let $f: R - \left\{-\frac{1}{2}\right\} \rightarrow R$ be defined by $f(x) = \frac{x-2}{2x+1}$. If $\alpha$ and $\beta$ satisfy the equation $f(f(x)) = -x$,then $4(\alpha^2 + \beta^2) = $

If $f(x)=e^x$ and $h(x)=(f \circ f)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module