Question

nc.m4.af.1 apply properties of function composition to build new functions from existing functions. clarification: students will extend their understanding of function notation and evaluating functions to working with compositions of two functions. this includes evaluating a composition of functions for specific values in the domain. students will understand that the net effect of f ∘ g is a → g(a) → f(g(a)) for any real value, a. students understand that a is in the domain of g and that g(a) is in the domain of f. there is no limit as to the type of functions to be used when composing functions. in problems involving roots in the denominator, it is not necessary to rationalize the denominator. checking for understanding indicator: find (f ∘ g)(x) if f(x)=-6x + 11 and g(x)=3x - 5. indicator: given the two functions, f(x)=2x - 3 and g(x)=x² + 1, find the following: a. f(g(x)) b. g(f(x)) indicator: given f(x)=\frac{1}{x^{2}+4} and g(x)=\sqrt{x + 1}. find f(g(x)) and g(f(x)). c. (f ∘ f)(x) d. (g ∘ g)(x)

Explanation:

Step1: Recall function - composition formula

The composition of functions $(f\circ g)(x)=f(g(x))$ means we substitute $g(x)$ into $f(x)$.

Step2: Solve for part a

Given $f(x)=2x - 3$ and $g(x)=x^{2}+1$. Then $f(g(x))=2(x^{2}+1)-3$. Expand the expression: $f(g(x))=2x^{2}+2 - 3=2x^{2}-1$.

Step3: Solve for part b

$g(f(x))=g(2x - 3)=(2x - 3)^{2}+1$. Expand $(2x - 3)^{2}$ using the formula $(a - b)^{2}=a^{2}-2ab + b^{2}$, where $a = 2x$ and $b = 3$. So $g(f(x))=(2x)^{2}-2\times(2x)\times3+3^{2}+1=4x^{2}-12x + 9 + 1=4x^{2}-12x+10$.

Step4: Solve for part c

$(f\circ f)(x)=f(f(x))$. Since $f(x)=2x - 3$, then $f(f(x))=2(2x - 3)-3$. Expand the expression: $f(f(x))=4x-6 - 3=4x - 9$.

Step5: Solve for part d

$(g\circ g)(x)=g(g(x))$. Since $g(x)=x^{2}+1$, then $g(g(x))=(x^{2}+1)^{2}+1$. Expand $(x^{2}+1)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, where $a=x^{2}$ and $b = 1$. So $g(g(x))=(x^{2})^{2}+2x^{2}+1 + 1=x^{4}+2x^{2}+2$.

Step6: Solve for the other composition

Given $f(x)=-6x + 11$ and $g(x)=3x - 5$. Then $(f\circ g)(x)=f(g(x))=-6(3x - 5)+11$. Expand the expression: $(f\circ g)(x)=-18x+30 + 11=-18x + 41$.

Answer:

a. $f(g(x))=2x^{2}-1$
b. $g(f(x))=4x^{2}-12x + 10$
c. $(f\circ f)(x)=4x - 9$
d. $(g\circ g)(x)=x^{4}+2x^{2}+2$
For $f(x)=-6x + 11$ and $g(x)=3x - 5$, $(f\circ g)(x)=-18x + 41$