Homework 2
Directions:
- Show each step of your work and fully simplify each expression.
- Turn in your answers in class on a physical piece of paper.
- Staple multiple sheets together.
- Feel free to use Desmos for graphing.
Answer the following:
- What is the definition of like terms?
- Expand and simplify each expression.
- $(2x^2 + 3x) + (3x^3 + 2x)$
- $(x+1)(x-2)$
- $(x^2 + 2x + 1)(x-2)$
- $(1 - x)^2$
- $(x^6 - x^5) -2(x^4 - x^3) - x(x^2 - x)$
- Factor the following expressions.
- $-2x^3 - x^2$
- $(x+3)^2(x-2) + (x+3)(x-2)^2$
- $x^2 + 5x + 6$
- $x^2 + 13x + 12$
- $2x^2 + 7x + 3$
- $2x^2yz + 7xyz + 3yz$
- $3x^4 - 16x^2 + 5$
- $(3x+2)^2 + 8(3x+2) + 12$
- Isolate $x$ in the following expressions:
- $3x + d = q$
- $ax + bx = c$
- $3x + b = 4x - 3$
- A student tries to simplify \[\dfrac{x^{-1} + y^{-1}}{4} = \dfrac{1}{4xy}\] What mistake did the student make?
- A student tries to simplify \[\dfrac{x+3}{x + 2} \cdot 4 = \dfrac{x+3\cdot 4}{x+2} = \dfrac{x+12}{x+2}\] What mistake did the student make?
- Show why $x = 2$ is a solution to the equation \[\dfrac{1}{x} - \dfrac{1}{x-4} = 1\]
- Find all real-valued solutions for the following equations.
- $3x + 4 = 7$
- $2x + 3 = 7 - 3x$
- $x^2 + x - 12 =0$
- $2x^2 = 8$
- $\dfrac{1}{x} = \dfrac{4}{3x} + 1$
- $\dfrac{1}{x-1} + \dfrac{1}{x+2} = \dfrac{5}{4}$
- Isolate $x$ in the equation $a(b + cx) + d = e$. Simplify any compound fractions.
- A student tries to isolate $x$ in the equation $(a + b)x = c + d$ by dividing by $a + b$, resulting in \[x = \dfrac{c}{a + b} + d\] What mistake did the student make?
- Suppose I have an expression $f(x)$ and I find two different inputs that give the same evaluation. In particular, I find that $x = 2$ and $x = 3$ gives $f(2) = f(3)$. Is $f$ a function?
- Suppose I have an expression $f(x)$ and I find one input which gives two different evaluations. In particular, I find that $x = 2$ spits out $f(2) = 5$ and $f(2) = 3$. Is $f$ a function?
- Write down two examples of functions in your daily life.
- Let $f(x) = x^2 - x + 1$. Evaluate the following:
- $f(1)$
- $f(a)$
- $f(-a)$
- $f(x + h)$
- Let \[f(x) = \begin{cases} 3x + 2 & x \geq 2 \\ x^2 - x & x < 2\end{cases}\] Evaluate the following:
- $f(0)$
- $f(1)$
- $f(2)$
- $f(3)$
- Suppose $f$ is a function. What two problems do you need to look for when finding the domain?
- Find the domain of the following functions:
- $f(x) = \dfrac{1}{x - 1} + \dfrac{1}{x - 2}$
- $f(x) = \sqrt{x - 1} + \dfrac{1}{x - 3}$
- Graph the points $(1, 2), (-1, 4), (-3,-2)$ and $(3, -4)$ in the coordinate plane.
- Sketch a graph of the function $f(x) = x^2-1$ drawn to scale on the coordinate plane.
- In the coordinate plane, what physical meaning does $f(x)$ take on?
- Draw the general shape of the functions $f(x) = x^2, g(x) = x^4, h(x) = x^6$ in the coordinate plane.
Do the same for the odd powers $f(x) = x^3, g(x) = x^5, h(x) = x^7$.
The rest of these problems will be on next week's homework. Skip for now.
- Draw a curve in the plane that is not a function.
- Given the functions $f(x) = \sqrt{x - 2}$ and $g(x) = 5 - x^2$, find the following:
- $f(g(0))$
- $f\circ g$
- $g \circ f$
- $f \circ f$
- $g \circ g$
- Given the function $F(x) = \dfrac{1}{\sqrt{x - 3}}$, write $F(x)$ as a composition of functions $f$ and $g$ (meaning determine $f$ and $g$ such that $F(x) = (f\circ g)(x)$).
- Suppose \[f(x) = x \qquad g(x) = \dfrac{1}{x-2} \qquad h(x) = \sqrt{x} \qquad i(x) = \dfrac{1}{x-3}\]
- Find $f(x) + g(x) - h(x) - i(x)$ and it's domain.
- Find $f(x) \cdot g(x) \cdot h(x) \cdot i(x)$ and it's domain.
- Suppose \[f(x) = x^2 \qquad g(x) = \sqrt{x} \qquad h(x) = \dfrac{1}{x - 1} \qquad i(x) = \dfrac{1}{x-2}\]
- Find $f(x) + g(x) - h(x) - i(x)$ and it's domain.
- Find $\dfrac{f(x) \cdot g(x) \cdot h(x)}{i(x)}$ and it's domain.