Respuesta :
Answer:
[tex](g \cdot h)(x)=x^2-2x+23[/tex]
Step-by-step explanation:
For composite functions, it's important to understand what the functions mean:
[tex](g\cdot h)(x)[/tex] which is read as "g of h, of x" means [tex]g ( \text{ }h(x) \text{ })[/tex] which is read as "g of, h of x" (with slight pauses at the comma). This means that x goes into the h function, and the output of the h function goes into the g function.
Putting "x" into the h function
[tex]h(x)=-x+4[/tex]
Since it is just "x" going into the h function, the function as written is the output when x is the input.
Putting the h function output, into the g function
[tex]g(x)=x^2-6x+3[/tex]
[tex]g(h(x))=(h(x))^2-6(h(x))+3[/tex]
Substitute
[tex]g(h(x))=(-x+4)^2+-6(-x+4)+3[/tex]
Squaring means the something multiplied by itself
[tex]g(h(x))=(-x+4)*(-x+4)+-6(-x+4)+3[/tex]
Use distributive property; (some people know binomial distribution as "FOIL" -- First, Outer, Inner, Last):
[tex]g(h(x))=[(-x)(-x)+4(-x)+4(-x)+4*4)]+[6x+4]+3[/tex]
Simplify the binomial terms:
[tex]g(h(x))=[x^2-8x+16]+[6x+4]+3[/tex]
Group like terms:
[tex]g(h(x))=x^2-2x+23[/tex]
Remember that [tex](g\cdot h)(x)[/tex] means [tex]g ( \text{ }h(x) \text{ })[/tex]
[tex](g \cdot h)(x)=x^2-2x+23[/tex]
So, [tex](g \cdot h)(x)=x^2-2x+23[/tex]
