What is the range of f(x)=x^2+9,x<= 0 ?

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range of f(x)=x^2+9,x<= 0

range

f (x) = x^{2} + 9, x \leq 0

f (x) \geq 9

Interval Notation

[9, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

x^{2} + 9, x \leq 0 : x \leq 0

Extreme Points of

x^{2} + 9, x \leq 0 :

Minimum

(0, 9)

Find the range for the interval

- \infty < x \leq 0 : 9 \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq 9

ln (x^{2} - 1)

f (x) = x + 8 - 1

\frac{2}{3} x - \frac{1}{3}

\frac{x ^{2} + x + 1}{x}

f (x) = x^{4} + 16 x^{3} + 54 x^{2} + 2

What is the range of f(x)=x^2+9,x<= 0 ?
The range of f(x)=x^2+9,x<= 0 is f(x)>= 9