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:
For each red function, write out what the transformation from $f(x)$ should be:
Graph the following functions using transformations. Include at least two transformed points and ordered of transformations you apply them in for full credit.
$g(x) = \sqrt{x + 4} - 3$
$g(x) = -(x-1)^2$
$g(x) = -2 - \sqrt{-2x + 6}$
$g(x) = -\lvert-2x - 2\rvert + 4$
$g(x) = 2(-2x+6)^2 - 1$
Suppose \[f(x) = 2x^2 + x \qquad g(x) = x^2 - x \qquad h(x) = x^2 - 1 \qquad k(x) = x^4\] Evaluate and expand the following:
$f(x) + g(x) + h(x)$
$f(x) - g(x)$ Hint Look at how many terms you are substituting. Don't forget...
$g(x)h(x) - k(x)$
$f(x)\left[k(x)\right]^2$
$\dfrac{g(x)}{h(x)}$ (also find the domain).
Consider $f(x) = x^2 -1$ and $g(x) = x^2 + 2x + 1$. We make a new function $h(x) = \dfrac{g(x)}{f(x)}$.
$\dfrac{g'(x)f(x) - f'(x)g(x)}{\left[g(x)\right]^2}$ Hint Look at how many terms you are substituting. Don't forget...
$f'(x)g(x) + f(x)g'(x)$
Given the functions $f(x) = 2x- 5$ and $g(x) = 3 - x^2$, evaluate the following and fully simplify.
Hint Expand because like terms are created.
$f\circ g$
$f(g(0))$
$g \circ f$
$f \circ f$
$g \circ g$
$f \circ f \circ f$
Given the following functions $F(x)$, find two functions $f$ and $g$ where $f\circ g = F$. Choosing $f(x) = x$ or $g(x) = x$ is zero credit for this problem.
$F(x) = (x-3)^4$
$F(x) = (2x-3)^{-\frac{1}{2}}$
$F(x) = \sqrt{1 - \sqrt{x}}$
$F(x) = \sqrt{(x^2 + 2x + 3)^3}$
Consider $y = f(x)$, where $x$ is time and $y$ is temperature. If you find the inverse $f^{-1}(x)$, what does the $x$ in the inverse mean?
Suppose $f$ is not a one-to-one function. Explain what property is violated when we try to define $f^{-1}$.
Are $f(x)$ and $g(x)$ inverses of each other? Show using the Inverse Function Property.
$f(x) = x^2, \qquad g(x) = \sqrt{x}$
$f(x) = 2 - 5x, \qquad g(x) = \dfrac{2 - x}{5}$
$f(x) = \dfrac{1}{x-1}, \qquad g(x) = \dfrac{1 + x}{x}$ Hint This is a compound fraction. I advise evaluating $f\circ g$ so you only need to deal with the denominator.
The function $f(x) = x^2$ is not one-to-one. What could we restrict the domain to to make $f(x)$ one-to-one? Hint Look at the graph. Which part can you delete?
Find the codomain of the following functions.
$f(x) = \sqrt{x}$
$f(x) = x^2 - 77$ Hint Transformations.
$f(x) = \lvert x \rvert + 1$
Find the inverse function of each of the following functions or explain why it is not possible.
Hint Check one-to-one first. Maybe inverse doesn't exist.
$f(x) = 3x + 5$
$f(x) = (x-1)^2$
$f(x) = \dfrac{x}{x + 2}$
$f(x) = 4-x^2, \qquad x \geq 0$
$f(x) = \sqrt{4 - x^2}, \qquad 0 \leq x \leq 2$
$f(x) = \lvert 3x - 6 \rvert, \qquad x \geq 0$
$f(x) = 2 + \sqrt{x + 3}$ Hint Similar technique to solving root equations.
$f(x) = \dfrac{2x + 4}{x - 7}$
Given this graph of a function:
find
$f^{-1}(2)$
$f^{-1}(5)$
$f^{-1}(6)$
Hint If $f(x) = y$, then $f^{-1}(y) = x$. What is the $y$?