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Popular >

range of f(x)=1+x^2

Solution

range

f (x) = 1 + x^{2}

Solution

f (x) \geq 1

+1

Solution steps

Vertex of

1 + x^{2} :

Minimum

(0, 1)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

If

a < 0

the range is

f (x) \leq y_{v}

If

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (0, 1)

f (x) \geq 1

Popular Examples

extreme f(x)=x+(25)/x extreme points

f (x) = x + \frac{25}{x}

inverse of f(x)=\sqrt[3]{x-9}+4 inverse

domain of f(x)=sqrt(36-x)domain

f (x) = 36 - x

periodicity of-6sin(3x+pi/2)periodicity

- 6 sin (3 x + \frac{π}{2})

vertices y=2x^2-2 vertices

y = 2 x^{2} - 2

Frequently Asked Questions (FAQ)

What is the range of f(x)=1+x^2 ?
The range of f(x)=1+x^2 is f(x)>= 1