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Popular >

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

Solution

range

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

Solution

f (x) \leq \frac{1}{2}

+1

Solution steps

Vertex of

- 2 x^{2} + 2 x :

Maximum

(\frac{1}{2}, \frac{1}{2})

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 = - 2,

Vertex

(x_{v}, y_{v}) = (\frac{1}{2}, \frac{1}{2})

f (x) \leq \frac{1}{2}

Popular Examples

inflection f(x)=x^{1/3}inflection points

f (x) = x^{\frac{1}{3}}

parity f(x)= 1/(x-1)parity

f (x) = \frac{1}{x - 1}

critical f(x)=(ln(x))/x critical points

f (x) = \frac{ln ( x )}{x}

domain of f(x)=1+sqrt(x)domain

f (x) = 1 + x

domain of f(x)=(x+6)/(24-sqrt(x^2-49))domain

f (x) = \frac{x + 6}{24 - x ^{2} - 49}

Frequently Asked Questions (FAQ)

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