What is the f(x)={2x+1:x>= 1,x^2-2:x<0} ?

The f(x)={2x+1:x>= 1,x^2-2:x<0} is

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f(x)={2x+1:x>= 1,x^2-2:x<0}

f (x) = {2 x + 1 : x \geq 1, x^{2} - 2 : x < 0}

Domain : x < 0 or x \geq 1

X Intercepts : (- 2, 0)

Asymptotes : Slant y = 2 x + 1

Monotone Intervals : Decreasing : - \infty < x < 0, Increasing : 1 < x < \infty

Interval Notation

Domain : (- \infty, 0) \cup [1, \infty)

Solution steps

Domain of

{2 x + 1 : x \geq 1, x^{2} - 2 : x < 0} : x < 0 or x \geq 1

Axis interception points of

{2 x + 1 : x \geq 1, x^{2} - 2 : x < 0} :

X Intercepts

: (- 2, 0)

Asymptotes of

{2 x + 1 : x \geq 1, x^{2} - 2 : x < 0} :

Slant

: y = 2 x + 1

Monotone Intervals of

{2 x + 1 : x \geq 1, x^{2} - 2 : x < 0} :

Decreasing

: - \infty < x < 0,

Increasing

: 1 < x < \infty

f (x) = 2 x^{4} - 8 x^{3} + 15 x^{2} + 120 x - 72

y = x + 11

y = ln (5) x^{5}

f (x) = 5 x^{2} - 2 x + 7

f (t) = (t^{2} - 4)^{\frac{2}{3}}