What is the range of g(x)=sin(x)+1 ?

The range of g(x)=sin(x)+1 is 0<= f(x)<= 2

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range of g(x)=sin(x)+1

range

g (x) = sin (x) + 1

0 \leq f (x) \leq 2

Interval Notation

[0, 2]

Solution steps

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

Add to the edges of the range:

1

0 \leq sin (x) + 1 \leq 2

Therefore the range is

0 \leq f (x) \leq 2

cr i t i c a l f (x) = (x + 4) (x - 1)^{2}

in v erse f (x) = x^{2} - 6 x + 13

a sy m pt o t es y = 2 \cdot 3^{x} - 3

in v erse y = 3 x - 2

d o main f (x) = x^{3} - 9 x