Prove that

Question:

Let $f: R \rightarrow R$ be defined by

$f(x)=\left\{\begin{array}{ccc}2 x+3, & \text { when } & x<-2 \\ x^{2}-2, & \text { when } & -2 \leq x \leq 3 \\ 3 x-1, & \text { when } & x>3\end{array}\right.$

Find (i) $f(2)$ (ii) $f(4)$ (iii) $f(-1)$ (iv) $f(-3)$.

 

 

Solution:

i) $f(2)$

Since $f(x)=x^{2}-2$, when $x=2$

$\therefore f(2)=(2)^{2}-2=4-2=2$

$\therefore f(2)=2$

ii)f(4)

Since $f(x)=3 x-1$, when $x=4$

$\therefore f(4)=(3 \times 4)-1=12-1=11$

$\therefore f(4)=11$

iii)f( - 1)

Since $f(x)=x^{2}-2$, when $x=-1$

$\therefore f(-1)=(-1)^{2}-2=1-2=-1$

$\therefore f(-1)=-1$

iv) $f(-3)$

Since $f(x)=2 x+3$, when $x=-3$

$\therefore f(-3)=2 \times(-3)+3=-6+3=-3$

$\therefore f(-3)=-3$

 

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