What is the range of sqrt(25-x^2),-5<= x<5 ?

The range of sqrt(25-x^2),-5<= x<5 is 0<= f(x)<= 5

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range of sqrt(25-x^2),-5<= x<5

range

25 - x^{2}, - 5 \leq x < 5

0 \leq f (x) \leq 5

Interval Notation

[0, 5]

Solution steps

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

Domain of

25 - x^{2}, - 5 \leq x < 5 : - 5 \leq x < 5

Extreme Points of

25 - x^{2}, - 5 \leq x < 5 :

Minimum

(- 5, 0),

Maximum

(0, 5)

Find the range for the interval

- 5 \leq x < 5 : 0 \leq f (x) \leq 5

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

0 \leq f (x) \leq 5

π + arcsin (2 x - 1)

f (x) = lo g_{2} (2 - ∣ 1 - x ∣)

tan (a r cos (\frac{2}{2}))

f (x) = - ∣ x - 3 ∣ + 2

What is the range of sqrt(25-x^2),-5<= x<5 ?
The range of sqrt(25-x^2),-5<= x<5 is 0<= f(x)<= 5