Question

Suppose a function

X

on

R

is given as

X

(

ω

) =

1

5

−

ω

2

, ω

≤

=

7

1 +

ω

2

, ω >

1. Identify the range of

X

.

2. Find

X

−

1

(

B

) when

B

= (0

,

1),

B

= [1

,

3],

B

= 4,

B

= (2

,

4)

Answer 1

Part 1

$\mathrm{The\:set\:of\:values\:of\:the\:dependent\:variable\:for\:which\:a\:function\:is\:defined}\\ \mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)\\ \mathrm{If}\:a<0\:\mathrm{the\:range\:is}\:f\left(x\right)\le \:y_v\\ a=-1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:15\right)\\ f\left(x\right)\le \:15\\ \mathrm{Range\:of\:}15-x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:15\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:15]\end{bmatrix}$

Part 2

The function is $71+w^2dw$

$X=71+w^2\\ \implies \mathrm{A\:function\:g\:is\:the\:inverse\:of\:function\:f\:if\:for}\:y=f\left(x\right),\:\:x=g\left(y\right)\:\\ X=71+w^2\\ w=71+X^2\\ X^{-1}=\sqrt{w-71}$