What is the f(x)=x^2-3x,g(x)=sqrt(x-1),g(x)\circ f(x) ?

The f(x)=x^2-3x,g(x)=sqrt(x-1),g(x)\circ f(x) is sqrt(x^2-3x-1)

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f(x)=x^2-3x,g(x)=sqrt(x-1),g(x)\circ f(x)

f (x) = x^{2} - 3 x, g (x) = x - 1, g (x) \circ f (x)

x^{2} - 3 x - 1

Solution steps

Find

g \circ f

given

f (x) = x^{2} - 3 x, g (x) = x - 1

g \circ f = g (f (x))

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

= g (x^{2} - 3 x)

g (x^{2} - 3 x) : x^{2} - 3 x - 1

= x^{2} - 3 x - 1

y = x - 4

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

f (x) = \frac{3 x + 5}{2 x - 3}

x^{4} - 12 x^{3} + 48 x^{2} - 64 x

y = 2 x

What is the f(x)=x^2-3x,g(x)=sqrt(x-1),g(x)\circ f(x) ?
The f(x)=x^2-3x,g(x)=sqrt(x-1),g(x)\circ f(x) is sqrt(x^2-3x-1)