Given f(x), find g(x) and h(x) such that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x f(x)= square root (-x+3)-1

f(x) = (-x + 3)^-1

g(x) = x^-1

h(x) = (-x + 3)

Step-by-step explanation:

We know that:

g(x) ≠ x

h(x) ≠ x

f(x) = g(h(x))

f(x) = (-x + 3)^(-1) = g(h(x))

Then we can define:

h(x) as the thing inside the first parenthesis:

h(x) = (-x + 3)

And g(x) as the function that is evaluated in this, then:

g(x) = x^-1

So when we have:

g(h(x)) = (h(x))^-1 = ( (-x + 3) )^-1 = (-x + 3)^-1

Then we have:

f(x) = (-x + 3)^-1

g(x) = x^-1

h(x) = (-x + 3)

step-by-step explanation:

y=mx+b

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)