What is the range of y=cos(4x)+1 ?

The range of y=cos(4x)+1 is 0<= f(x)<= 2

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range of y=cos(4x)+1

range

y = cos (4 x) + 1

0 \leq f (x) \leq 2

Interval Notation

[0, 2]

Solution steps

The range of the basic

cos

function is

- 1 \leq cos (4 x) \leq 1

- 1 \leq cos (4 x) \leq 1

Add to the edges of the range:

1

0 \leq cos (4 x) + 1 \leq 2

Therefore the range is

0 \leq f (x) \leq 2

f (x) = 6^{x} + 3

f (x) = 2 - 2^{(a t a n ((x - 1)^{2}))}

f (x) = \frac{1 + 2 ^{x}}{4 - 2 ^{x}}

f (x) = \frac{x}{x - 2}

(- \frac{7}{3}, \frac{1}{3}), (- \frac{5}{3}, - \frac{7}{3})