We did a similar thing in completing the square (66
➤
).
Themodulus functionf(x)=|x|is defined by (8
➤
)
|x|=xifx≥ 0
=−xifx< 0Its graph is shown in Figure 3.4 and it is clearly even.
xyy = x0Figure 3.4The modulus functiony=|x|.
Solution to review question 3.1.4(i) For f(x)= 3 x^3 −xwe have
f(−x)= 3 (−x)^3 −(−x)=− 3 x^3 +x
=−( 3 x^3 −x)=−f(x)
So 3x^3 −xis an odd function.(ii) For f(x)=x^2
1 +x^2f(−x)=(−x)^2
1 +(−x)^2=x^2
1 +x^2=f(x)so this is even.(iii) Iff(x)=2 x
x^2 − 1thenf(−x)=2 (−x)
(−x)^2 − 1=−2 x
x^2 − 1=−f(x)sof(x)is odd.(iv) Iff(x)=x^2
x+ 1we havef(−x)=(−x)^2
−x+ 1=x^2
1 −x
This is not equal tof(x)or−f(x)and so this function is neither
odd nor even.