Question : Quadratics and Functions
(a) Express \(3x^{2}-12x+14\) in the form \(3(x+a)^{2}+b\) where \(a\) and \(b\) are constants to be found.
The function \(f(x)=3x^{2}-12x+14\) is defined for \(x \geq k\) where \(k\) is a constant.
(b) Find the least value of \(k\) for which the function \(f^{-1}\) exists.
For the rest of this question, you should assume that \(k\) has the value found in part (b).
(c) Find an expression for \(f^{-1}(x)\).
(d) Hence or otherwise solve the equation \(ff(x)=29\).
Working space:
▶️Answer/Explanation
Answer: \(3(x – 2)^2 + 2\)
Working:
Rewrite by completing the square:
\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]
So, \(a = -2\), \(b = 2\).
Key Concept:
Completing the square transforms a quadratic expression into vertex form, revealing the vertex and constants \(a\) and \(b\).
Working space:
▶️Answer/Explanation
Answer: \(k = 2\)
Working:
\(f(x) = 3(x – 2)^2 + 2\) has a vertex at \(x = 2\). For \(f^{-1}\) to exist, \(f\) must be one-to-one. Since it’s increasing for \(x \geq 2\), the smallest \(k\) is 2.
Key Concept:
A function has an inverse if it is one-to-one, achieved by restricting the domain to where the function is strictly increasing or decreasing.
Working space:
▶️Answer/Explanation
Answer: \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\)
Working:
With domain \(x \geq 2\), set \(y = 3(x – 2)^2 + 2\):
\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]
So, \(f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}\), for \(x \geq 2\).
Key Concept:
The inverse function is found by solving \(y = f(x)\) for \(x\), ensuring the domain restriction maintains one-to-one behavior.
Working space:
▶️Answer/Explanation
Answer: \(x = 3\)
Working:
\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]
\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]
Since \(x \geq 2\), \(x = 3\).
Key Concept:
Composite functions and domain restrictions are used to solve equations involving function iterations.
Syllabus Reference
Quadratics and Functions
- (a) SL 1.1 – Quadratic functions: completing the square
- (b) SL 1.2 – Functions: conditions for existence of inverse
- (c) SL 1.2 – Functions: finding inverse functions
- (d) SL 1.2 – Functions: composite functions and solving equations
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question : Functions
In the diagram, the graph with equation \( y=f(x) \) is shown with solid lines and the graph with equation \( y=g(x) \) is shown with broken lines.
Working space:
▶️Answer/Explanation
Answer:
1. Reflection in y-axis
2. Translation or shift \( \begin{pmatrix} -1 \\ 0 \end{pmatrix} \)
3. Stretch, factor 2, parallel to y-axis
Working:
Correct order and three correctly named transformations only.
Alternative Solution for first 3 marks:
1. Translation or shift \( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)
2. Reflection in y-axis
Alternative solutions:
There are alternative solutions which can be marked in the same way, e.g., the given stretch, translation \( \begin{pmatrix} -4 \\ 0 \end{pmatrix} \), reflect in \( x=-2.5 \).
Working space:
▶️Answer/Explanation
Answer: \( g(x) = 2f(-x-1) \) or \( a=2 \), \( b=-1 \), \( c=-1 \)
Syllabus Reference
Functions
- (a) SL 1.2 – Transformations of functions
- (b) SL 1.2 – Composite transformations and function notation
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question : Functions
The function \( f \) is defined by \( f(x) = \frac{2x + 1}{2x – 1} \) for \( x < \frac{1}{2} \).
The function \( g \) is defined by \( g(x) = 3x + 2 \) for \( x \in \mathbb{R} \).
Working space:
▶️Answer/Explanation
Correct answer: \( \frac{1}{3} \)
Working:
\( f(-1) = \frac{2(-1) + 1}{2(-1) – 1} = \frac{-2 + 1}{-2 – 1} = \frac{-1}{-3} = \frac{1}{3} \)
Key Concept:
Evaluate a function by substituting the given value into the expression.
Working space:
▶️Answer/Explanation
Correct answer:
Working:
The graph of \( y = f^{-1}(x) \) is the reflection of \( y = f(x) \) over the line \( y = x \). The mirror line \( y = x \) is shown on the graph.
Key Concept:
The inverse function’s graph is obtained by reflecting the original function’s graph over the line \( y = x \).
Working space:
▶️Answer/Explanation
Correct answer: \( f^{-1}(x) = \frac{x + 1}{2(x – 1)} \), Domain: \( x < 1 \)
Working:
Let \( y = f(x) = \frac{2x + 1}{2x – 1} \).
Solve for \( x \):
\[ y = \frac{2x + 1}{2x – 1} \]
\[ 2x + 1 = y(2x – 1) \]
\[ 2x + 1 = 2xy – y \]
\[ 2xy – 2x = y + 1 \]
\[ 2x(y – 1) = y + 1 \]
\[ x = \frac{y + 1}{2(y – 1)} \]
Thus, \( f^{-1}(x) = \frac{x + 1}{2(x – 1)} \).
Alternative form: \( f^{-1}(x) = \frac{-x – 1}{2 – 2x} \).
Domain of \( f^{-1} \): The denominator \( 2(x – 1) \neq 0 \), so \( x \neq 1 \). Since the range of \( f(x) \) for \( x < \frac{1}{2} \) is \( y < 1 \), the domain of \( f^{-1} \) is \( x < 1 \).
Alternative Method:
\[ y = \frac{2x + 1}{2x – 1} = 1 + \frac{2}{2x – 1} \]
\[ y – 1 = \frac{2}{2x – 1} \]
\[ 2x – 1 = \frac{2}{y – 1} \]
\[ 2x = \frac{2}{y – 1} + 1 \]
This leads to the same expression after simplification.
Key Concept:
The inverse function is found by swapping \( x \) and \( y \) and solving for \( y \). The domain of \( f^{-1} \) is the range of \( f \).
Working space:
▶️Answer/Explanation
Correct answer: \( x = \frac{3}{8} \)
Working:
First, compute \( f\left(\frac{1}{4}\right) \):
\[ f\left(\frac{1}{4}\right) = \frac{2 \cdot \frac{1}{4} + 1}{2 \cdot \frac{1}{4} – 1} = \frac{\frac{1}{2} + 1}{\frac{1}{2} – 1} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3 \]
Then, compute \( g(f(\frac{1}{4})) \):
\[ g(-3) = 3(-3) + 2 = -9 + 2 = -7 \]
The equation is:
\[ f(x) = g(f(\frac{1}{4})) = -7 \]
\[ \frac{2x + 1}{2x – 1} = -7 \]
Solve:
\[ 2x + 1 = -7(2x – 1) \]
\[ 2x + 1 = -14x + 7 \]
\[ 16x = 6 \]
\[ x = \frac{6}{16} = \frac{3}{8} \]
Alternative solution:
Since \( f(x) = -7 \), use the inverse:
\[ x = f^{-1}(-7) \]
\[ f^{-1}(x) = \frac{x + 1}{2(x – 1)} \]
\[ f^{-1}(-7) = \frac{-7 + 1}{2(-7 – 1)} = \frac{-6}{-16} = \frac{3}{8} \]
Key Concept:
Solve composite function equations by evaluating the inner function first, then solving the resulting equation.
Syllabus Reference
Functions
- (a)(i) ALV: 1.2 – Evaluating functions
- (a)(ii) ALV: 1.2 – Graphing inverse functions
- (a)(iii) ALV: 1.2 – Finding the inverse function and its domain
- (b) ALV: 1.2 – Solving equations involving composite functions
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)
Question : Functions
The function \( f \) is defined by \( f(x) = 3 + 6x – 2x^2 \) for \( x \in \mathbb{R} \).
Working space:
▶️Answer/Explanation
Correct answer: \( f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2 \), range \( f(x) \leq \frac{15}{2} \)
Working:
Rewrite \( f(x) = 3 + 6x – 2x^2 \) by completing the square:
\[ f(x) = -2x^2 + 6x + 3 \]
Factor out \(-2\) from the \( x \) terms:
\[ f(x) = -2(x^2 – 3x) + 3 \]
Complete the square:
\[ x^2 – 3x = \left(x – \frac{3}{2}\right)^2 – \frac{9}{4} \]
So:
\[ f(x) = -2\left[\left(x – \frac{3}{2}\right)^2 – \frac{9}{4}\right] + 3 \]
\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{18}{4} + 3 \]
\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{18}{4} + \frac{12}{4} \]
\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{30}{4} \]
\[ = -2\left(x – \frac{3}{2}\right)^2 + \frac{15}{2} \]
Thus, \( a = \frac{15}{2} \), \( b = 2 \), \( c = \frac{3}{2} \).
Range: Since \(-2\left(x – \frac{3}{2}\right)^2 \leq 0\), the maximum occurs at \( x = \frac{3}{2} \):
\[ f\left(\frac{3}{2}\right) = \frac{15}{2} \]
As \( x \to \pm \infty \), \( f(x) \to -\infty \). So, the range is:
\[ f(x) \leq \frac{15}{2} \]
Key Concept:
Completing the square transforms a quadratic into vertex form, revealing the maximum/minimum and range.
Working space:
▶️Answer/Explanation
Correct answer: Reflect in the \( x \)-axis, then translate left by \( \frac{3}{2} \) and up by \( \frac{15}{2} \).
Working:
\( f(x) = \frac{15}{2} – 2\left(x – \frac{3}{2}\right)^2 \) has a maximum at \( \left(\frac{3}{2}, \frac{15}{2}\right) \). We need \( h(x) \) with a minimum at \( (0, 0) \).
1. Reflection: Reflect in the \( x \)-axis:
\[ y = -f(x) = -\left[3 + 6x – 2x^2\right] = 2x^2 – 6x – 3 \]
Vertex becomes \( \left(\frac{3}{2}, -\frac{15}{2}\right) \), a minimum.
2. Translation: Shift from \( \left(\frac{3}{2}, -\frac{15}{2}\right) \) to \( (0, 0) \):
- Left by \( \frac{3}{2} \): replace \( x \) with \( x + \frac{3}{2} \)
- Up by \( \frac{15}{2} \): add \( \frac{15}{2} \)
\[ h(x) = 2\left(x + \frac{3}{2}\right)^2 – 6\left(x + \frac{3}{2}\right) – 3 + \frac{15}{2} \]
Simplify:
\[ = 2\left(x^2 + 3x + \frac{9}{4}\right) – 6x – 9 – 3 + \frac{15}{2} \]
\[ = 2x^2 + 6x + \frac{18}{4} – 6x – 12 + \frac{15}{2} = 2x^2 + \frac{9}{2} – 12 + \frac{15}{2} = 2x^2 \]
So, \( h(x) = 2x^2 \), minimum at \( (0, 0) \).
Key Concept:
Transformations like reflections and translations adjust the position and orientation of a graph’s vertex.
Working space:
▶️Answer/Explanation
Correct answer: \( g(x) \) sketch as parabola; not one-to-one unless restricted.
Working:
Assuming \( g(x) = f(x) = 3 + 6x – 2x^2 \):
– Sketch: Downward parabola, vertex at \( \left(\frac{3}{2}, \frac{15}{2}\right) \).
– One-to-one: \( f(x) \) is not one-to-one (parabola fails horizontal line test) unless restricted.
Key Concept:
A function is one-to-one if each output corresponds to exactly one input, often requiring domain restriction for quadratics.
Working space:
▶️Answer/Explanation
Correct answer: \( g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2} \), sketch with \( y = x \) mirror line.
Working:
Restrict \( f(x) \) to \( x \geq \frac{3}{2} \) (decreasing branch):
\[ y = 3 + 6x – 2x^2 \]
Solve for \( x \):
\[ 2x^2 – 6x + (3 – y) = 0 \]
Discriminant: \( \Delta = (-6)^2 – 4 \cdot 2 \cdot (3 – y) = 36 – 24 + 8y = 12 + 8y \)
\[ x = \frac{6 \pm \sqrt{12 + 8y}}{4} \]
Take \( + \) for \( x \geq \frac{3}{2} \):
\[ x = \frac{6 + \sqrt{12 + 8y}}{4} = \frac{6 + 2\sqrt{3 + 2y}}{4} = \frac{3 + \sqrt{3 + 2y}}{2} \]
\[ g^{-1}(x) = \frac{3 + \sqrt{3 + 2x}}{2}, \quad x \leq \frac{15}{2} \]
Sketch: \( y = g(x) \) (right branch), \( y = g^{-1}(x) \) reflected over \( y = x \).
Key Concept:
The inverse function’s graph is a reflection over \( y = x \), requiring a one-to-one function via domain restriction.
Syllabus Reference
Functions
- (a) Topic-1.2 – Functions
- (b) Topic-1.2 – Functions
- (c) Topic-1.2 – Functions
- (d) Topic-1.2 – Functions
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)