# **Mathematics 105 Exam 3 Practice (old)**
**November, 2024**

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> **Answer the questions clearly.**
> No calculators, books or notes. Show all work.

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### 1. Solve the equation:
$$2(2x+1)-3(-x+5)=4(3x-1)$$

### Solution:

{%youtube gYsXD1OwdtU %}

\begin{align}
2(2x+1)-3(-x+5)&=4(3x-1) \\
(2)(2x)+(2)(1)+(-3)(-x)+(-3)(5)&=(4)(3x)+(4)(-1) \\
4x+2+3x-15&=12x-4 \\
7x-13&=12x-4 \\
-5x-13&=-4 \\
-5x&=9 \\
x&=-\dfrac{9}{5}
\end{align}

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### 2. Let $f(x)=x^4$ and $g(x)=2x^5-3x^2+4$. Calculate and simplify:

- $(fg)(x)=$

{%youtube gYsXD1OwdtU %}
### Solution:
\begin{align}
(x^4)(2x^5-3x^2+4)&=(x^4)(2x^5)+(x^4)(-3x^2)+(x^4)(4) \\
&=2x^9-3x^6+4x^4
\end{align}

- $(g/f)(x)=$

### Solution:

\begin{align}
\dfrac{2x^5-3x^2+4}{x^4}&=\dfrac{2x^5}{x^4}-\dfrac{3x^2}{x^4}+\dfrac{4}{x^4} \\
&=2x^{5-4}-3x^{2-4}+4x^{0-4} \\
&=2x^{1}-3x^{-2}+4x^{-4} \\
&=2x-3x^{-2}+4x^{-4}
\end{align}

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### 3. Let $f(x)=2x^5$ and $g(x)=3x^4-2x^3+2$. Calculate and simplify:

- $(g/f)(x)=$

- $(f/g)(x)=$

### Solution:

- $(g/f)(x)=\dfrac{3x^4-2x^3+2}{2x^5}=\dfrac{3x^4}{2x^5}-\dfrac{2x^3}{2x^5}+\dfrac{2}{2x^5}=\dfrac{3}{2}x^{-1}-x^{-2}+x^{-5}$

- $(f/g)(x)=\dfrac{2x^5}{3x^4-2x^3+2}$

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### 4. Let $f(x)=x^2+2$ and $g(x)=3x-2$. Calculate and simplify:

- $(f \circ g)(x)=$

- $(g \circ f)(x)=$

### Solution:

- $(f \circ g)(x)$ means plug $g$ into $f$.

$(f \circ g)(x)=(3x-2)^2+2$

Simplification:

\begin{align}
(f \circ g)(x)&=(3x-2)^2+2 \\
&=(3x-2)(3x-2)+2 \\
&=(3x)(3x)+(3x)(-2)+(-2)(3x)+(-2)(-2)+2 \\
&=9x^2-6x-6x+4+2 \\
&=9x^2-12x+6
\end{align}

- $(g \circ f)(x)$ means plug $f$ into $g$.

$(g \circ f)(x)=3(x^2+2)-2$

Simplification:

\begin{align}
(g \circ f)(x)&=3(x^2+2)-2 \\
&=(3)(x^2)+(3)(2)-2 \\
&=3x^2+6-2 \\
&=3x^2+4
\end{align}

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### 5. Let $h(x)=(7-2x)^4$. Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)=$

- $g(x)=$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=u^4$|$g(x)=7-2x$ |

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### 6. Let
$$h(x)=\frac{1}{(2x+5)^3}$$
Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)$

- $g(x)$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=\dfrac{1}{u^3}$|$g(x)=2x+5$ |

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### 7. Let
$$h(x)=\sqrt{\frac{2x+1}{3x-5}}$$
Find $f(x)$ and $g(x)$ such that $h(x)=(f \circ g)(x)$.

- $f(x)=$

- $g(x)=$

### Solution:

| Outside Function| Inside Function|
|---|---|
|$f(u)=\sqrt{u}$|$g(x)=\dfrac{2x+1}{3x-5}$ |

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### 8. Solve the equation:
$(3x+4)(x-2)=0$

### Solution:

$$(3x+4)(x-2)=0$$

Zero Product Property
|$3x+4=0$| $x-2=0$|
|---|---|
|$3x=-4$| $x=2$|
|$x=-\dfrac{4}{3}$|$x=2$|

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### 9. Solve the equation:
$3x^2-15=0$

### Solution:

\begin{align}
3x^2-15&=0 \\
\dfrac{3x^2-15}{3}&=\dfrac{0}{3} \\
\dfrac{3x^2}{3}-\dfrac{15}{3}&=0 \\
x^2-5&=0 \\
x^2&=5 \\
\sqrt{x^2}&=\pm \sqrt{5} \\
x&=\pm \sqrt{5} \\
x&=-\sqrt{5},x=+\sqrt{5}
\end{align}

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### 10. Solve the equation:
$7x^2=4x$

### Solution:

\begin{align}
7x^2&=4x \\
7x^2-4x&=0 \\
x(7x)+x(-4)&=0 \\
x(7x-4)&=0
\end{align}

Zero Product Property

|$x=0$|$7x-4=0$|
|---|---|
|$x=0$|$7x=4$|
|$x=0$|$x=\dfrac{4}{7}$|

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### 11. Solve the equation by factoring:
$x^2-10x-24=0$

### Solution:

$x^2$ splits as $(x)(x)$.

$-24$ splits as:
1. $(24)(-1)$,
2. $(-24)(1)$
3. $(2)(-12)$
4. $(-2)(12)$
5. $(3)(-8)$
6. $(-3)(8)$
7. $(4)(-6)$
8. $(-4)(6)$

Setting up all the possibilities and FOILing out to check which one works:

|Possibilities|FOIL work| FOIL simplified|
|---|---|---|
|$(x+24)(x-1)$| $x^2-x+24x-24$| $x^2+23x-24$|
|$(x-24)(x+1)$| $x^2+x-24x-24$| $x^2-23x-24$|
|$\checkmark$ $(x+2)(x-12)$ $\checkmark$| $x^2-12x+2x-24$| $\checkmark$$x^2-10x-24$ $\checkmark$|
|$(x-2)(x+12)$| $x^2+12x-2x-24$| $x^2+10x-24$|
|$(x+3)(x-8)$| $x^2-8x+3x-24$| $x^2-5x-24$|
|$(x-3)(x+8)$| $x^2+8x-3x-24$| $x^2+5x-24$|
|$(x+4)(x-6)$| $x^2-6x+4x-24$| $x^2-2x-24$|
|$(x-4)(x+6)$| $x^2+6x-4x-24$| $x^2+2x-24$|

Thus we factor the equation as

$$(x+2)(x-12)=0$$

Zero Product Property

|$x+2=0$|$x-12=0$|
|---|---|
|$x=-2$|$x=12$|

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### 12. Solve the equation by factoring:
$6x^2-2x=4$

### Solution:

Get it into $ax^2+bx+c=0$ form by subtracting 4 from both sides to get:

$$6x^2-2x-4=0$$

Then dividing both sides by the greatest common factor GCF $2$:

$$3x^2-x-2=0$$

$3x^2$ splits as $(3x)(x)$.

$-2$ splits as:
1. $(2)(-1)$,
2. $(-2)(1)$
3. $(1)(-2)$
4. $(-1)(2)$

Setting up all the possibilities and FOILing out to check which one works:

|Possibilities|FOIL work| FOIL simplified|
|---|---|---|
| $\checkmark$ $(3x+2)(x-1)$ $\checkmark$ | $3x^2-3x+2x-2$ | $\checkmark$ $3x^2-x-2$ $\checkmark$|
|$(3x-2)(x+1)$ | $3x^2+3x-2x-2$ | $3x^2+x-2$ |
|$(3x+1)(x-2)$ | $3x^2-6x+x-2$ | $3x^2-5x-2$ |
|$(3x-1)(x+2)$ | $3x^2+6x-x-2$ | $3x^2+5x-2$ |

Thus we factor the equation as

$$(3x+2)(x-1)=0$$

Zero Product Property

|$3x+2=0$|$x-1=0$|
|---|---|
|$3x=-2$|$x=1$|
|$x=-\dfrac{2}{3}$|$x=1$|

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### 13. Solve the equation with the quadratic formula:
$$2x^2-2x-5=0$$
Use:
$$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

### Solution:

#### Step 1. Make sure it's in $ax^2+bx+c=0$ form:

$$2x^2-2x-5=0$$ is in the form $ax^2+bx+c=0$ $\checkmark$

#### Step 2. Identify the $a,b,c$ as the numbers in front of each term (coefficients).
$a=2$
$b=-2$
$c=-5$

#### Step 3. Calculate $b^2-4ac$:

\begin{align}
b^2-4ac&=(2)^2-4(2)(-5) \\
&=4-4(2)(-5) \\
&=4+(-4)(2)(-5) \\
&=4+(-8)(-5) \\
&=4+40 \\
&=44
\end{align}

#### Step 4. Square root and simplify.

$$\sqrt{44}=\sqrt{4}\sqrt{11}=2\sqrt{11}$$

#### Step 5. Plug it into the last part of the Quadratic formula:

\begin{align}
x&=\dfrac{-b \pm 2\sqrt{11}}{2a} \\
x&=\dfrac{-(-2) \pm 2\sqrt{11}}{2(2)} \\
x&=\dfrac{2 \pm 2\sqrt{11}}{4} \\
x&=\dfrac{2}{4} \pm \dfrac{2\sqrt{11}}{4} \\
x&=\dfrac{1}{2} \pm \dfrac{\sqrt{11}}{2} \\
\end{align}

$x=\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}$
and $x=\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}$.

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### 14. Solve the equation:
$x^4-8x^2+7=0$

### Solution:

Substitute $u=x^2$ into $x^4-8x^2+7=0$:

\begin{align}
x^4-8x^2+7&=0 \\
u^2-8u+7&=0 \\
(u-1)(u-7)&=0
\end{align}

Zero Product Property

|$u-1=0$|$u-7=0$|
|---|---|
|$u=1$|$u=7$|
|$x^2=1$|$x^2=7$|
|$\sqrt{x^2}=\pm \sqrt{1}$ | $\sqrt{x^2}=\pm \sqrt{7}$|
|$x=\pm 1$| $x=\pm \sqrt{7}$|

Four solutions:

$x=-1$, $x=1$, $x=-\sqrt{7}$, and $x=\sqrt{7}$.

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### 15. Solve the equation:
$y^3+7y^2+12y=0$

### Solution:

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### 16. Solve:
$$\frac{1}{4}+\frac{4}{x}=\frac{1}{3}+\frac{3}{x}$$

### Solution:

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### 17. Solve:
$$7+\sqrt{2x-5}=10$$

### Solution:

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### 18. For (a) and (b), circle one of the four sketches to describe the end behavior of the graph of the function.
**Circle the leading term of the polynomial.**

**(a)** $f(x) = -3x^3 - 2x^2 - x + 4$
(Circle leading term above.)

*(Insert sketches)*

### Solution:

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**(b)** $g(x) = -2x^3 + 4x^4 + 20x$
(Circle leading term above.)

*(Insert sketches)*

### Solution:

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### 19. Find the zeroes of
$$k(x) = (x+1)^3(x+4)^2x^5$$
and state the multiplicity of each.

### Solution:

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### 20. Let
$$k(x) = -2x^4(x+2)$$

**(a)** Find the zeroes and state the multiplicity of each.

**(b)** What is the leading term of $k(x)$?

**(c)** Draw a rough graph, illustrating end behavior and behavior at intercepts (cross or bounce).

*(Insert graph grid)*

### Solution:

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### 21. Let
$$g(x)=\frac{3x^2+2}{x^2+5x}$$
Find the vertical asymptotes and horizontal asymptotes of $g(x)$.

### Solution:

#### Vertical Asymptote: Set bottom=0

\begin{align}
x^2+5x&=0 \\
x(x)+x(5)&=0 \\
x(x+5)&=0
\end{align}

Zero Product Property

|$x=0$|$x+5=0$|
|---|---|
|$x=0$|$x=-5$|

$g(x)$ has two vertical asymptotes: $x=0$ and $x=-5$.

#### Horizontal Asymptote: Ignore lower terms

$$g(x)=\frac{3x^2+2}{x^2+5x}$$

Ignore the lower terms 2 and 5x:

$$\frac{3x^2+2}{x^2+5x}=\dfrac{3x^2+2}{x^2+5x} \approx \dfrac{3x^2}{x^2}=3$$

$g(x)$ has a horizontal asymptote $y=3$.

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### 22. Let
$$f(x)=\frac{2x+2}{3x-6}$$
Graph $y = f(x)$, labeling:

- **x-intercepts:**

- **y-intercept:**

- **Vertical asymptotes:**

- **Horizontal asymptotes:**

*(Insert graph grid)*

### Solution:

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