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

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

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