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for questions 10-12, given the equations: $f(x) = x^2 - 3x - 15$ and $g…

Question

for questions 10-12, given the equations: $f(x) = x^2 - 3x - 15$ and $g(x) = 2x - 4$, answer the following. 7 pts 10. $(f + g)(x)$ 11. $(f - g)(3)$ 12. $(f * g)(2)$

Explanation:

Question 10:

Step1: Recall the definition of function addition

The sum of two functions \((f + g)(x)\) is defined as \(f(x) + g(x)\).

Step2: Substitute the given functions

Given \(f(x)=x^{2}-3x - 15\) and \(g(x)=2x - 4\), we substitute these into the formula:

$$ LATEXBLOCK0 $$

Step3: Combine like terms

Combine the \(x\)-terms and the constant terms:

$$ LATEXBLOCK1 $$

Step1: Recall the definition of function subtraction

The difference of two functions \((f - g)(x)\) is defined as \(f(x)-g(x)\). To find \((f - g)(3)\), we first find the expression for \((f - g)(x)\) and then substitute \(x = 3\).

Step2: Find \((f - g)(x)\)

Substitute \(f(x)=x^{2}-3x - 15\) and \(g(x)=2x - 4\) into \(f(x)-g(x)\):

$$ LATEXBLOCK0 $$

Step3: Simplify \((f - g)(x)\)

Distribute the negative sign and combine like terms:

$$ LATEXBLOCK1 $$

Step4: Substitute \(x = 3\) into \((f - g)(x)\)

Now substitute \(x = 3\) into \(x^{2}-5x - 11\):

$$ LATEXBLOCK2 $$

Step5: Calculate the result

$$ 9 - 15 - 11=-17 $$

Step1: Recall the definition of function multiplication

The product of two functions \((f\cdot g)(x)\) is defined as \(f(x)\cdot g(x)\). To find \((f\cdot g)(2)\), we first find the expression for \((f\cdot g)(x)\) and then substitute \(x = 2\).

Step2: Find \((f\cdot g)(x)\)

Substitute \(f(x)=x^{2}-3x - 15\) and \(g(x)=2x - 4\) into \(f(x)\cdot g(x)\):

$$ LATEXBLOCK0 $$

Step3: Expand the product

Use the distributive property (FOIL method for binomials, extended for polynomials):

$$ LATEXBLOCK1 $$

Step4: Combine like terms

$$ LATEXBLOCK2 $$

Step5: Substitute \(x = 2\) into \((f\cdot g)(x)\)

Substitute \(x = 2\) into \(2x^{3}-10x^{2}-18x + 60\):

$$ LATEXBLOCK3 $$

Step6: Calculate the result

$$ 16-40-36 + 60=0 $$

Answer:

\((f + g)(x)=x^{2}-x - 19\)

Question 11: