What is the range of f(x)=(-x^2+6x-8)^{(1/2)} ?

The range of f(x)=(-x^2+6x-8)^{(1/2)} is 0<= f(x)<= 1

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range of f(x)=(-x^2+6x-8)^{(1/2)}

range

f (x) = (- x^{2} + 6 x - 8)^{(\frac{1}{2})}

0 \leq f (x) \leq 1

Interval Notation

[0, 1]

Solution steps

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

Domain of

(- x^{2} + 6 x - 8)^{\frac{1}{2}} : 2 \leq x \leq 4

Extreme Points of

(- x^{2} + 6 x - 8)^{\frac{1}{2}} :

Minimum

(2, 0),

Maximum

(3, 1),

Minimum

(4, 0)

Find the range for the interval

2 \leq x \leq 4 : 0 \leq f (x) \leq 1

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

0 \leq f (x) \leq 1

f (x) = x - 3 + 4

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

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

\frac{2}{( 2 x - 5 )}

f (x) = 2 x^{2} + x - 3

What is the range of f(x)=(-x^2+6x-8)^{(1/2)} ?
The range of f(x)=(-x^2+6x-8)^{(1/2)} is 0<= f(x)<= 1