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f(x)=7x+2 g(x)=(x^(2))/(x-5) Write (g@f)(x) as an expression in terms of x. (g@f)(x)=

f(x)=7x+2 f(x)=7 x+2 \newlineg(x)=x2x−5 g(x)=\frac{x^{2}}{x-5} \newlineWrite (g∘f)(x) (g \circ f)(x) as an expression in terms of x x .\newline(g∘f)(x)= (g \circ f)(x)=

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Q. f(x)=7x+2 f(x)=7 x+2 \newlineg(x)=x2x−5 g(x)=\frac{x^{2}}{x-5} \newlineWrite (g∘f)(x) (g \circ f)(x) as an expression in terms of x x .\newline(g∘f)(x)= (g \circ f)(x)=
  1. Understand (g@f)(x)(g@f)(x): First, let's understand what (g@f)(x)(g@f)(x) means. It means we need to apply the function gg to the result of the function ff applied to xx. In other words, we need to substitute f(x)f(x) into g(x)g(x) wherever we see an xx in g(x)g(x).
  2. Write Functions: Now let's write down the functions f(x)f(x) and g(x)g(x) to see what we're working with:\newlinef(x)=7x+2f(x) = 7x + 2\newlineg(x)=x2x−5g(x) = \frac{x^2}{x - 5}
  3. Substitute f(x)f(x) into g(x)g(x): We need to substitute f(x)f(x) into g(x)g(x). This means wherever we see an xx in g(x)g(x), we replace it with 7x+27x + 2:
    (g@f)(x)=(7x+2)2(7x+2)−5(g@f)(x) = \frac{(7x + 2)^2}{(7x + 2) - 5}
  4. Simplify the Expression: Now let's simplify the expression:\newline(g@f)(x)=(7x+2)2(7x+2−5)(g@f)(x) = \frac{(7x + 2)^2}{(7x + 2 - 5)}\newline(g@f)(x)=(7x+2)2(7x−3)(g@f)(x) = \frac{(7x + 2)^2}{(7x - 3)}
  5. Express in Terms of x: We have now expressed (g@f)(x)(g@f)(x) in terms of xx. There is no further simplification needed unless we want to expand the numerator, which is not required by the question prompt.

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