If #f(x) = x^2-3x-23# and #g(x) = x-1#, how do you find #(f*g)(x)#?

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Answer 1 · 2017-04-24T20:19:18

See below.

#(f*g)(x)# is the same as #f(x)*g(x)#.

Then, this is just:

#(x^2-3x-23)(x-1)=x^3-x^2-3x^2+3x-23x+23#

Combining like terms,

#=x^3-4x^2-20x+23#

Answer 2 · 2017-04-24T20:38:10

#(f*g)(x)=x^2-5x-19#

When you read #(f*g)(x)# read it as #"g"# inside of #"f"#.
This means we are taking the value of #"g"# and putting it into #"f"#

If they told us to solve #(g*f)(x)# it would mean #"f"# inside of #"g"# and it would look like this:
#(g*f)(x)=(x^2-3x-23)-1#

We have:

#(f*g)(x)=(x-1)^2-3(x-1)-23#

Exponents first

#(f*g)(x)=x^2-2x+1-3(x-1)-23#

Now multiply

#(f*g)(x)=x^2-2x+1-3x+3-23#

Organize / Add and subtract common variables

#(f*g)(x)=x^2-5x-19#