What is the range of f(x)=(x+1\circ x^2-1) ?

The range of f(x)=(x+1\circ x^2-1) is f(x)>=-5/4

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range of f(x)=(x+1\circ x^2-1)

range

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

f (x) \geq - \frac{5}{4}

Interval Notation

[- \frac{5}{4}, \infty)

Solution steps

Vertex of

x + 1 \circ x^{2} - 1 :

Minimum

(- \frac{1}{2 \circ}, \frac{1}{4 \circ ^{2}} - \frac{1}{2 \circ} - 1)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

a < 0

the range is

f (x) \leq y_{v}

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (- \frac{1}{2}, - \frac{5}{4})

f (x) \geq - \frac{5}{4}

- 3 x^{4} + 24 x^{3} - 48 x^{2}

f (x) = - 4 x

f (x) = \frac{1}{x ^{2} - 9}

h (x) = 2 x - 6

4 x - 3 y = 4

What is the range of f(x)=(x+1\circ x^2-1) ?
The range of f(x)=(x+1\circ x^2-1) is f(x)>=-5/4