Question: Transformation of Functions: Oct 26
Name: Elsie Mzogwe
Big Idea 1: students have a conceptual understanding of mathematical situations
1. Given the graph of $y = f(x)$, identify any potential invariant point(s) under the transformation
$g(x)= -2f(x)$?
y = f(x)
invariant point $(-1,0)$
2. The graph of $y = f(x)$ is transformed into the graph $g(x) = -\frac{1}{2} f\left(\frac{1}{2} x + 2\right) - 1$. The
ordered pair $(4, 6)$ is found on the point of the original function $y = f(x)$. Explain
how you know the location of $(4, 6)$ after the function has been transformed and
identify its new location.
3. A function $f(x) = x^2 - x - 2$ is multiplied by a constant value $k$ to create a new function
g(x) = kf(x)$. If the graph of $y = g(x)$ passes through point $(3, 14)$, state the value
of $k$.
Graph description: In the upper-left graph, the piecewise graph of $y = f(x)$ has an open circle near $(-4, 2)$, a closed point at $(-3, -2)$, a closed point at $(-1, 0)$, a closed point at $(1, 1)$, and a horizontal segment from $(1, 1)$ to $(3, 1)$ with an open circle at $(3, 1)$; position_hint = top-left.
4. Answer the following questions with respect to the graph that is found below:
y = f(x)
a) Without sketching, identify the domain and range of the inverse.
$D(f) = \{4 \geq y \geq -2, y \in \mathbb{R}\}$
$R(f) = \{-6 \leq x \leq 2, x \in \mathbb{R}\}$
b) Will the inverse be a function? If not, restrict the domain on $f(x)$ so that the inverse is a function.
Big Idea 2: students apply my mathematical knowledge to solve mathematical situations
1. Below is the function $y = \sqrt{x - 2}$.
$(2, 0)$
$(3, 1)$
$(6, 2)$
On the graph below, draw the inverse function. State the equation of the inverse functions.
Draw the inverse function and state the equation of this inverse. Include any restrictions that have been used so that the it looks like the function drawn below.
graph top-left: piecewise graph with an open circle at $(-5, 4)$ connected by a horizontal segment to a closed circle at $(-2, 4)$, then a line segment down to a closed point at $(0, -2)$, then a line segment up to a closed point at $(2, 3)$. position_hint=top-left
2. Given the function $f(x) = 3x + 5$, if $x = -25$ on the inverse function, explain how
you would determine the $y$ value on the inverse function. Why is this your method
of choice. NOTE: You do NOT need to solve this question mathematically.
3. Given the graph $y = f(x)$, sketch the graph of $y - 1 = 3f\left(- \frac{1}{2}(x - 4)\right)$
position_hint: top-left, a line graph for $y = f(x)$ is shown with points approximately at $(-3,-1)$, $(-2,1)$, $(2,3)$, and $(4,-3)$, connected by straight segments; position_hint: bottom-right, a transformed line graph is shown with corresponding points approximately at $(-1,-10)$, $(0,8)$, $(4,2)$, and $(6,-4)$, connected by straight segments.
1. **Problem 1: Identify invariant points under $g(x) = -2f(x)$ given $y = f(x)$ and invariant point $(-1,0)$.**
- An invariant point $(a,b)$ satisfies $g(a) = b$.
- Since $g(x) = -2f(x)$, for the point $(-1,0)$, check if $g(-1) = 0$.
- Given $f(-1) = 0$ (from the invariant point), then $g(-1) = -2 \times 0 = 0$.
- So $(-1,0)$ remains unchanged under the transformation and is an invariant point.
2. **Problem 2: Find the new location of point $(4,6)$ under $g(x) = -\frac{1}{2} f\left(\frac{1}{2} x + 2\right) - 1$.**
- Original point: $(4,6)$ means $f(4) = 6$.
- To find the new $x$ coordinate $x_{new}$ such that $g(x_{new}) = y_{new}$ corresponds to $f(4)$, solve:
$$\frac{1}{2} x_{new} + 2 = 4$$
- Solve for $x_{new}$:
$$\frac{1}{2} x_{new} = 4 - 2 = 2$$
$$x_{new} = 4$$
- Now find $y_{new}$:
$$g(4) = -\frac{1}{2} f(4) - 1 = -\frac{1}{2} \times 6 - 1 = -3 - 1 = -4$$
- So the new location of the point is $(4, -4)$.
3. **Problem 3: Find $k$ if $g(x) = kf(x)$ passes through $(3,14)$ and $f(x) = x^2 - x - 2$.**
- Calculate $f(3)$:
$$f(3) = 3^2 - 3 - 2 = 9 - 3 - 2 = 4$$
- Since $g(3) = k f(3) = 14$, solve for $k$:
$$k \times 4 = 14$$
$$k = \frac{14}{4} = \frac{7}{2} = 3.5$$
4. **Problem 4a: Identify domain and range of the inverse of $y = f(x)$ given:**
- Domain of $f$: $D(f) = \{4 \geq y \geq -2, y \in \mathbb{R}\}$
- Range of $f$: $R(f) = \{-6 \leq x \leq 2, x \in \mathbb{R}\}$
- The domain of the inverse function $f^{-1}$ is the range of $f$.
- The range of the inverse function $f^{-1}$ is the domain of $f$.
- Therefore:
$$D(f^{-1}) = R(f) = \{-6 \leq x \leq 2\}$$
$$R(f^{-1}) = D(f) = \{4 \geq y \geq -2\}$$
5. **Problem 4b: Will the inverse be a function? If not, restrict domain of $f(x)$ so inverse is a function.**
- The inverse is a function if $f$ is one-to-one (passes horizontal line test).
- If $f$ is not one-to-one on its entire domain, restrict domain to an interval where $f$ is one-to-one.
- For example, restrict domain to an interval where $f$ is strictly increasing or decreasing.
6. **Problem 5: Find inverse of $y = \sqrt{x - 2}$ and state restrictions.**
- Start with:
$$y = \sqrt{x - 2}$$
- Swap $x$ and $y$ to find inverse:
$$x = \sqrt{y - 2}$$
- Square both sides:
$$x^2 = y - 2$$
- Solve for $y$:
$$y = x^2 + 2$$
- Domain of original function: $x \geq 2$ (since inside square root must be $\geq 0$).
- Range of original function: $y \geq 0$.
- Therefore, domain of inverse is $x \geq 0$ and range is $y \geq 2$.
- **Inverse function:**
$$f^{-1}(x) = x^2 + 2, \quad x \geq 0$$
7. **Problem 6: Given $f(x) = 3x + 5$, if $x = -25$ on inverse function, explain how to find $y$ on inverse.**
- The inverse function reverses $f$.
- To find $y = f^{-1}(-25)$, solve $f(y) = -25$.
- This method is chosen because the inverse function value at $x$ is the $y$ such that $f(y) = x$.
8. **Problem 7: Sketch graph of $y - 1 = 3f\left(- \frac{1}{2}(x - 4)\right)$.**
- This transformation involves:
- Horizontal shift by $4$ (inside function argument).
- Horizontal reflection and stretch by factor $2$ (due to $-\frac{1}{2}$ inside).
- Vertical stretch by $3$ and vertical shift up by $1$.
- Points transform as:
$$x_{new} = -2(x - 4)$$
$$y_{new} = 3 f(x_{new}) + 1$$
- Use given points to plot transformed graph accordingly.
**Final answers:**
- Problem 1 invariant point: $(-1,0)$
- Problem 2 new point: $(4, -4)$
- Problem 3 value of $k$: $3.5$
- Problem 4a inverse domain and range: $D(f^{-1}) = [-6, 2]$, $R(f^{-1}) = [-2, 4]$
- Problem 4b inverse is function if domain restricted to one-to-one interval
- Problem 5 inverse function: $f^{-1}(x) = x^2 + 2$, $x \geq 0$
- Problem 6 method: solve $f(y) = x$ to find inverse value
- Problem 7 transformation explained for sketching