Question

Determine whether each function is even, odd, or neither.
f(x) = x²_9

g(x) = |x-3

f(x) = 22-1

g(x) = x + x2

Answer 1

Rule for even

$\\ \sf\longmapsto \boxed{\sf f(-x)=f(x),x\in D_f}$

Rule for odd

$\\ \sf\longmapsto \boxed{\sf f(-x)=-f(x),D_f}$

• Here D_f means domain of function.

#1

$\\ \sf\longmapsto f(x)=\sqrt{x^2}-9$

• Take x=4

$\\ \sf\longmapsto f(4)=\sqrt{4^2}-9=4-9=-5$

$\\ \sf\longmapsto f(-4)=\sqrt{(-4)^2}=4-9=-5$

Even function✓

#2

$\\ \sf\longmapsto g(x)=|x-3|$

• Take x=2

$\\ \sf\longmapsto g(2)=|2-3|=|-1|=1$

$\\ \sf\longmapsto g(-2)=|-2-3|=|-5|=5$

$\\ \sf\longmapsto -g(2)=-1$

Odd function✓

#3

$\\ \sf\longmapsto f(x)=\dfrac{x}{x^2-1}$

• Take x=3

$\\ \sf\longmapsto f(3)=\dfrac{3}{3^2-1}=\dfrac{3}{9-1}=\dfrac{3}{8}$

$\\ \sf\longmapsto f(-3)=\dfrac{-3}{(-3)^2-1}=\dfrac{-3}{8}$

$\\ \sf\longmapsto -f(3)=\dfrac{-3}{8}$

Odd function ✓

#4

$\\ \sf\longmapsto g(x)=x+x^2$

• Take x=1

$\\ \sf\longmapsto g(1)=1+(1)^2=2$

$\\ \sf\longmapsto g(-1)=-1+(-1)^2=0$

$\\ \sf\longmapsto -g(1)=-2$

Neither✓