Question

Can somebody help me fix my 4 and 5 problem of this exercise?

Answer 1

Hello

To solve this question, we were given a particular function and asked to evalute when the function is defined with a particular variable

$\begin{gathered} f(x)=3-4x \\ g(x)=3x+4x^2 \end{gathered}$

For f(-6)

$\begin{gathered} f(x)=3-4x^{} \\ f(-6)=3-4(-6) \\ f(-6)=3-4(-6) \\ f(-6)=27 \end{gathered}$

For g(-9)

$\begin{gathered} g(x)=3x-4x^2 \\ g(-9)=3(-9)+4(-9)^2 \\ g(-9)=297 \end{gathered}$

For f(-9) + g(-9)

$\begin{gathered} f(x)=3-4x \\ g(x)=3x+4x^2 \\ f(-9)+g(-9)=3-4(-9)+3(-9)+4(-9)^2=336 \end{gathered}$

for g(-6) - f(-9)

$\begin{gathered} g(x)=3x+4x^2 \\ f(x)=3-4x \\ g(-6)-f(-9)=3(-6)+4(-6)^2-(3-4(-9))=87 \end{gathered}$

For f(-9).g(-7)

$\begin{gathered} g(x)=3x+4x^2 \\ f(x)=3-4x \\ f(-9).\text{g}(-7)=3-4(-9)\times3(-7)+4(-7)^2=39\times175=6825 \end{gathered}$

For f(-6) / g(-7)

$\begin{gathered} f(x)=3-4x \\ g(x)=3x+4x^2 \\ \frac{f(-6)}{g(-7)}=\frac{3-4(-6)}{3(-7)+4(-7)^2}=\frac{27}{175} \end{gathered}$

From the calculations above, i believe you must've seen your mistaken and taken the necessary correction.

Therefore the answers are

1 = 27

2 = 297

3 = 336

4 = 87

5 = 6825

6 = 27/175