MA 3972-MA-Book May 8, 2018 13:46Review of Precalculus 63- Solve fory.Thus,y=
x^2 + 1
2
.
- Replaceybyf−^1 (x). You have f−^1 (x)=
x^2 + 1
2
.
- Since the range off(x)is[0,∞), the domain of f−^1 (x)is[0,∞).
- Verifyf−^1 (x) by checking:
Sincex>0,
√
x^2 =x,f(f−^1 (x))=f(
x^2 + 1
2)
=√
2(
x^2 + 1
2)
− 1 =xf−^1 (f(x))=f−^1 (√
2 x−1)=(
√
2 x−1)^2 + 1
2
=xTrigonometric and Inverse Trigonometric Functions
There are six basic trigonometric functions and six inverse trigonometric functions. Their
graphs are illustrated in Figures 5.3-9 to 5.3-20.− 2− 10y = sin xDomain: {−∞ < x < ∞}
Range: {− 1 ≤ y ≤ 1}
Frequency: 1Amplitude: 1
Period: 2πxy− 2 π −1. 5π− 1 π −0.5π 0.5π 1 π 1. 5π 2 π12Figure 5.3-9xy1 2y = sin−^1 x− 2 − 1 00.5π−0.5π
Domain: {− 1 ≤ x ≤ 1}
Range: {− 2 π ≤ y ≤ 2 π}
Figure 5.3-10− 2− 10y = cos xxy− 2 π−1. 5π − 1 π −0.5π 0.5π 1 π 1. 5π 2 π
12Domain: {−∞ < x < ∞}
Range: {− 1 ≤ y ≤ 1}
Frequency: 1Amplitude: 1
Period: 2πFigure 5.3-11xy1 2y = cos–^1 x- 2 – 1 0
π0.5πDomain: {– 1 ≤ x ≤ 1}
Range: {0 ≤ y ≤ π}
Figure 5.3-12