What is the range of f(x)= 4/(sqrt(x^2-4x+3)) ?

The range of f(x)= 4/(sqrt(x^2-4x+3)) is f(x)>0

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range of f(x)= 4/(sqrt(x^2-4x+3))

range

f (x) = \frac{4}{x ^{2} - 4 x + 3}

f (x) > 0

Solution steps

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

Domain of

\frac{4}{x ^{2} - 4 x + 3} : x < 1 or x > 3

Extreme Points of

\frac{4}{x ^{2} - 4 x + 3} :

None

Find the range for the interval

- \infty < x < 1 : 0 < f (x) < \infty

Find the range for the interval

3 < x < \infty : 0 < f (x) < \infty

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

f (x) > 0 or f (x) > 0

f (x) > 0

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

8 x^{4} - 8 x^{2} + 1

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

f (x) = 6 x + 4

\frac{4 x ^{2} + 6 x - 4}{2 x ^{2} + 13 x + 15}

What is the range of f(x)= 4/(sqrt(x^2-4x+3)) ?
The range of f(x)= 4/(sqrt(x^2-4x+3)) is f(x)>0