What is the range of sqrt(49-x^2) ?

The range of sqrt(49-x^2) is 0<= f(x)<= 7

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range of sqrt(49-x^2)

range

49 - x^{2}

0 \leq f (x) \leq 7

Interval Notation

[0, 7]

Solution steps

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

Domain of

49 - x^{2} : - 7 \leq x \leq 7

Extreme Points of

49 - x^{2} :

Minimum

(- 7, 0),

Maximum

(0, 7),

Minimum

(7, 0)

Find the range for the interval

- 7 \leq x \leq 7 : 0 \leq f (x) \leq 7

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

0 \leq f (x) \leq 7

f (x) = 80 - 4.9 t^{2}

\frac{4 x ^{2} + 1}{2 x ^{2} + 5 x - 3}

y = 4 cos (π x + \frac{π}{2})

\frac{x + 2}{x}

y = (x - 1)^{2} + 2