What is the range of f(x)=sqrt(-x^2+6x-8) ?

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

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range of f(x)=sqrt(-x^2+6x-8)

range

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

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 : 2 \leq x \leq 4

Extreme Points of

- x^{2} + 6 x - 8 :

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

2 - x

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

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

f (x) = x^{2} - 12 x + 5

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

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