Math 1250 HW2 Notes: Inverse Functions & Parabolas
Inverse Functions
For the given functions, determine the inverse (where possible) and state the domain and range of each inverse.
a) f(x) = 5x + 4
Step 1: Replace f(x) with y
y = 5x + 4Step 2: Swap x and y
x = 5y + 4Step 3: Solve for y
Subtract 4 from both sides: x - 4 = 5y
Divide by 5: y = \frac{x - 4}{5}Step 4: Replace y with f^{-1}(x).
Inverse: f^{-1}(x) = \frac{x-4}{5}Step 5: Determine the domain and range of the inverse.
Since the inverse is a linear function, its domain is all real numbers.
Domain of inverse: (-\infty, \infty)
Since the inverse is a linear function, its range is all real numbers.
Range of inverse: (-\infty, \infty)
b) g(x) = (x-3)^2, \; x \le 3
Step 1: Replace g(x) with y
y = (x-3)^2Step 2: Swap x and y
x = (y-3)^2Step 3: Solve for y
Take the square root of both sides: \pm\sqrt{x} = y-3
Add 3 to both sides: y = 3 \pm\sqrt{x}
To determine whether to use the + or - sign, we consider the domain of the original function x \le 3. This means the output of the inverse function (which is the input of the original function) must be y \le 3. To get values less than or equal to 3, we must use the minus sign with the square root. For example, if x=1, 3+\sqrt{1}=4 (which is not \le 3) and 3-\sqrt{1}=2 (which is \le 3).
So, Inverse: g^{-1}(x) = 3 - \sqrt{x}Step 4: Determine the domain and range of the inverse.
The domain of the inverse is the range of the original function. For g(x) = (x-3)^2 with x \le 3, the vertex is at (3, 0). Since the parabola opens upwards and we are considering x \le 3, the minimum value of g(x) is 0, and it increases as x approaches - \infty. So, the range of g(x) is [0, \infty). This becomes the domain of the inverse.
Domain of inverse: [0, \infty) (since the original outputs are (y \ge 0)). Also, for \sqrt{x} to be defined, x \ge 0.
The range of the inverse is the domain of the original function. The original domain was x \le 3.
Range of inverse: (-\infty, 3] (the inverse maps back to the original domain x \le 3)
c) h(x) = \sqrt{4 - x^2}
Step 1: Determine if the function is one-to-one.
The natural domain of h(x) = \sqrt{4 - x^2} is [-2, 2] because 4 - x^2 \ge 0 \Rightarrow x^2 \le 4 \Rightarrow -2 \le x \le 2. This function represents the upper semicircle of a circle centered at the origin with radius 2. For example, h(-1) = \sqrt{3} and h(1) = \sqrt{3}. Since different inputs can give the same output, the function is not one-to-one on its natural domain [-2, 2], so it does not have an inverse without restricting the domain.Step 2: Restrict the domain to make it one-to-one.
If we restrict the domain to a half-interval, e.g. x \in [0, 2], the function becomes one-to-one.Step 3: Find the inverse for the restricted domain.
Let y = \sqrt{4 - x^2}. Swap x and y: x = \sqrt{4 - y^2}.
Square both sides: x^2 = 4 - y^2
Solve for y^2: y^2 = 4 - x^2
Take the square root: y = \pm\sqrt{4 - x^2}
For the restricted domain x \in [0, 2] (for the original function), the original outputs h(x) are values in [0, 2]. Since y represents the input of the original function ([0, 2]), and y must be positive in this interval, we choose the positive root.
Inverse: h^{-1}(x) = \sqrt{4 - x^2}Step 4: Determine the domain and range of the inverse for the restricted domain.
The domain of the inverse is the range of the restricted original function. For h(x) = \sqrt{4 - x^2} with x \in [0, 2], the minimum value is h(2) = \sqrt{0} = 0 and the maximum value is h(0) = \sqrt{4} = 2. So the range of h(x) is [0, 2].
Domain of the inverse: [0, 2]
The range of the inverse is the domain of the restricted original function, which was [0, 2].
Range of the inverse: [0, 2]Note on alternative restriction:
(If the original domain was restricted to x \in [-2, 0], the inverse would be h^{-1}(x) = -\sqrt{4 - x^2} to ensure the range of the inverse (original domain) is negative, with domain [0, 2] and range [-2, 0].)
Quadratic in Vertex Form
Given the quadratic f(x) = 2x^2 + 8x + 4, write in vertex form.
Step 1: Factor out the leading coefficient from the x^2 and x terms.
f(x) = 2(x^2 + 4x) + 4Step 2: Complete the square inside the parenthesis.
Take half of the coefficient of x (which is 4), square it ((\frac{4}{2})^2 = 2^2 = 4), then add and subtract it inside the parenthesis.
f(x) = 2(x^2 + 4x + 4 - 4) + 4Step 3: Rewrite the perfect square trinomial as a squared term.
f(x) = 2((x+2)^2 - 4) + 4Step 4: Distribute the factored-out coefficient and simplify.
f(x) = 2(x+2)^2 - 2(4) + 4
f(x) = 2(x+2)^2 - 8 + 4
f(x) = 2(x+2)^2 - 4Step 5: Identify the vertex.
From the standard vertex form a(x-h)^2 + k, the vertex is (h, k).
Comparing 2(x+2)^2 - 4 to a(x-h)^2 + k, we have h = -2 and k = -4.
Vertex: (-2, -4)
Parabola: Vertex Form from Vertex and a Point
Find the equation of a parabola with vertex at (-3, 2) that passes through (0, -1).
Step 1: Write down the general vertex form of a parabola.
y = a(x - h)^2 + kStep 2: Substitute the coordinates of the given vertex (h, k) into the vertex form.
Given vertex (-3, 2), so h = -3 and k = 2.
y = a(x - (-3))^2 + 2
Vertex form: y = a(x + 3)^2 + 2Step 3: Use the given point (x, y) to solve for the constant a.
The parabola passes through (0, -1), so substitute x = 0 and y = -1 into the equation from Step 2.
-1 = a(0 + 3)^2 + 2
-1 = a(3)^2 + 2
-1 = 9a + 2Step 4: Solve the equation for a
Subtract 2 from both sides: -1 - 2 = 9a \Rightarrow -3 = 9a
Divide by 9: a = \frac{-3}{9} = -\frac{1}{3}Step 5: Write the final equation by substituting the value of a back into the vertex form.
Equation: y = -\tfrac{1}{3}(x+3)^2 + 2
The Arch of a Bridge (Parabolic Arch)
An arch is modeled by a parabola with height 14 ft and base width 20 ft.
a) Height as a function of horizontal position x (with center at x = 0 and base endpoints at x = \pm 10):
Step 1: Identify key points and the vertex form.
The arch has a maximum height of 14 ft at its center, so the vertex is at (0, 14).
The total base width is 20 ft, and with the center at x=0, the base endpoints are at x = -10 and x = 10. At these points, the height (y-value) is 0.
We use the vertex form h(x) = a(x - h)^2 + k. With vertex (0, 14), this becomes h(x) = a(x - 0)^2 + 14, or h(x) = a x^2 + 14Step 2: Use a known point on the parabola to find the value of a.
We know that h(10) = 0 (or h(-10) = 0). Substitute x = 10 and h(x) = 0 into the equation from Step 1.
0 = a(10)^2 + 14
0 = 100a + 14
Subtract 14 from both sides: -14 = 100a
Divide by 100: a = -\frac{14}{100} = -\frac{7}{50}.Step 3: Write the final equation for the height.
Therefore: \boxed{h(x) = 14 - \frac{7}{50}x^2, \; -10 \le x \le 10}
b) Can an 8 ft wide, 12 ft tall truck pass?
Step 1: Determine the x-coordinates the truck occupies.
An 8 ft wide truck centered under the arch (x=0) occupies the interval from x = -4 to x = 4 (since half of the width is 4 ft).Step 2: Calculate the minimum height of the arch over the truck's width.
The height of the arch is lowest at the edges of the truck's width, i.e., at x = -4 or x = 4.
Substitute x=4 into the equation found in part (a): h(4) = 14 - \frac{7}{50}(4)^2
h(4) = 14 - \frac{7}{50}(16)
h(4) = 14 - \frac{112}{50}
To subtract, find a common denominator: h(4) = \frac{14 \times 50}{50} - \frac{112}{50} = \frac{700}{50} - \frac{112}{50} = \frac{588}{50} = \frac{294}{25}
Convert to decimal for comparison: \frac{294}{25} = 11.76\text{ ft}Step 3: Compare the available clearance to the truck's height.
The available clearance at the edges of the 8 ft wide truck is approximately 11.76 ft.
Since 11.76 ft < 12 ft, a truck of height 12 ft cannot pass.
c) Tallest 8 ft wide truck that can pass?
Step 1: Recognize that the limiting height is the minimum height over the truck's width.
As calculated in part (b), the minimum height available for an 8 ft wide truck is at its edges (x = \pm 4).Step 2: State the maximum height based on this calculation.
The limiting height is the height at the truckās side positions: h(4) = \frac{294}{25} \approx 11.76\text{ ft}.Step 3: Formulate the answer.
Therefore, the tallest 8-ft-wide truck that can pass has height \boxed{\frac{294}{25} \text{ ft} \approx 11.76 \text{ ft}}.
Graphs and Zeros: Lowest-Degree Equation
Write an equation for each function with the lowest possible degree that satisfies the following graphs (has the same zeros). Leave your answer in factored form.
A & B: The zeros (x-intercepts) must be read from the graphs. Since the graphs are not provided here, explicit equations cannot be written.
General method for writing an equation from zeros:
Step 1: Identify all zeros (x-intercepts) from the graph. Let these be r1, r2, \dots, r_n.
Step 2: Determine the multiplicity of each zero.
If the graph crosses the x-axis at a zero, its multiplicity (m_i) is odd (usually 1 for lowest degree).
If the graph touches the x-axis and turns around at a zero, its multiplicity (m_i) is even (usually 2 for lowest degree).
Step 3: Form the factors of the polynomial. For each zero ri with multiplicity mi, the corresponding factor is (x - ri)^{mi}.
Step 4: Assemble the polynomial in factored form. The equation will be the product of these factors, multiplied by a constant c.
P(x) = c\prodi (x - ri)^{mi} where \prodi denotes the product of all factors. The constant c is a nonzero constant (often assumed to be 1 if a specific leading coefficient or another point on the graph is not provided). If a specific point other than an x-intercept is given, you can substitute its coordinates into the equation to solve for c.
Note: If you provide the zeros observed from the graphs for A and B, I can convert them to explicit factored forms right away.