What is the range of f(x)=xsqrt(4-x^2) ?

The range of f(x)=xsqrt(4-x^2) is -2<= f(x)<= 2

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range of f(x)=xsqrt(4-x^2)

range

f (x) = x 4 - x^{2}

- 2 \leq f (x) \leq 2

Solution steps

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

Domain of

x 4 - x^{2} : - 2 \leq x \leq 2

Extreme Points of

x 4 - x^{2} :

Maximum

(- 2, 0),

Minimum

(- 2, - 2),

Maximum

(2, 2),

Minimum

(2, 0)

Find the range for the interval

- 2 \leq x \leq 2 : - 2 \leq f (x) \leq 2

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

- 2 \leq f (x) \leq 2

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

f (x) = 15 - x

x^{3} - 64

f (x) = lo g_{3} (- x - 4) - 1

f (x) = \frac{x ^{2}}{x ^{2} - 25}

What is the range of f(x)=xsqrt(4-x^2) ?
The range of f(x)=xsqrt(4-x^2) is -2<= f(x)<= 2