1) Introduction to Advanced Functions 🤔
By now, you’ve explored linear, quadratic, exponential, and logarithmic functions, plus some transformations (shifts, stretches, reflections). At Level 4, we delve deeper into:
- Function composition and its properties.
- Inverses of more complex functions (including domain/range restrictions).
- Piecewise & parametric forms for modeling more intricate behaviors.
- Advanced transformations beyond simple shifts and scales (reflections across lines, horizontal/vertical compressions).
- Additional types: absolute value and rational transformations, etc.
We’ll start simply to refresh, then progress to powerful transformations and compositions!
2) Composition of Functions 🏗️
2.1 Basic Composition
If \( f(x) \) and \( g(x) \) are functions, the composition \( (f\circ g)(x) \) means \( f(g(x)) \).
Example 1:
- Let \( f(x)=2x-1 \) and \( g(x)=x^2 \).
- \( (f\circ g)(x)= f(x^2)= 2x^2-1 \).
- \( (g\circ f)(x)= g(2x-1)= (2x-1)^2= 4x^2-4x+1 \).
Note: Order matters in composition.
2.2 Composing More Complex Functions
Composition may involve exponentials, logarithms, or piecewise definitions. Always check the domain.
Example 2 (Advanced):
- \( f(x)=\ln(x) \) and \( g(x)= e^x-2 \).
- \( (f\circ g)(x)= \ln(e^x-2) \), with domain \( e^x-2>0 \) (i.e. \( x>\ln2 \)).
3) Inverse Functions and Restrictions 🔄
An inverse function \( f^{-1} \) undoes the work of \( f \). For an invertible function, \( (f\circ f^{-1})(x)=x \).
3.1 Finding Inverses
- Write \( y=f(x) \).
- Solve for \( x \) in terms of \( y \).
- Swap variables to get \( f^{-1}(x) \).
Example 3: For \( f(x)=\frac{3x-4}{2} \):
- Let \( y=\frac{3x-4}{2} \).
- Solve: \( 2y=3x-4 \) so \( 3x=2y+4 \) and \( x=\frac{2y+4}{3} \).
- Thus, \( f^{-1}(x)=\frac{2x+4}{3} \).
3.2 Domain & Range
For an inverse to exist, \( f \) must be one-to-one. We often restrict the domain if needed.
Example 4: \( f(x)=x^2 \) is not invertible on all \( \mathbb{R} \). Restrict \( x\ge0 \) to define \( f^{-1}(x)=\sqrt{x} \).
4) Piecewise & Parametric Forms 📐
4.1 Piecewise-Defined Functions
A function may have different formulas on different intervals.
Example 5 (Piecewise):
\( f(x)= \begin{cases} x+2, & x<1 \\ x^2-1, & x\ge1 \end{cases} \)
Graph each section and check continuity at the transition point \( x=1 \).
4.2 Parametric Equations
Instead of expressing \( y \) directly as a function of \( x \), we define both in terms of a parameter \( t \).
Example 6:
\( x=2\cos t,\quad y=2\sin t,\quad t\in[0,2\pi] \)
This traces a circle of radius 2.
5) Advanced Transformations 🔧
5.1 Horizontal/Vertical Shifts & Scales
- \( y=f(x-h)+k \): shift right \( h \) and up \( k \).
- \( y=A\cdot f(x) \): vertical scaling by \(|A|\); if \(A<0\), reflect across the x-axis.
- \( y=f(Bx) \): horizontal scaling by \( 1/|B| \); if \(B<0\), reflect across the y-axis.
Example 7: If \( f(x)=x^3 \), then for \( g(x)=-2f(3x+1)+4 \):
- \( 3x+1 \) implies a horizontal shift left by \(\frac{1}{3}\) and a compression by factor \(\frac{1}{3}\).
- Multiplying by \(-2\) reflects vertically and scales by 2.
- Add 4 shifts the graph upward by 4.
5.2 Reflections Across Other Lines
Advanced transformations may reflect across lines (e.g., reflecting across \( y=x \) swaps the roles of \( x \) and \( y \), akin to finding the inverse).
6) Examples (Detailed) 🍀
Example 8 (Composition with Domain Constraints)
Problem: Let \( f(x)=\sqrt{x-1} \) and \( g(x)=(x+2)^2 \). Find \( (f\circ g)(x) \) and its domain.
- \( (f\circ g)(x)= \sqrt{(x+2)^2-1} \).
- Simplify: \(\sqrt{x^2+4x+3}\).
- Domain: Require \( x^2+4x+3\ge0 \). Factor as \((x+1)(x+3)\ge0\), so \( x\in(-\infty,-3]\cup[-1,\infty) \).
Example 9 (Inverse Function)
Problem: If \( f(x)=2e^x-1 \), find \( f^{-1}(x) \).
- Write \( y=2e^x-1 \) and solve: \( y+1=2e^x \) so \( e^x=\frac{y+1}{2} \).
- Take logs: \( x=\ln\Bigl(\frac{y+1}{2}\Bigr) \).
- Swap variables: \( f^{-1}(x)=\ln\Bigl(\frac{x+1}{2}\Bigr) \). (Domain for inverse: \( x>-1 \)).
Example 10 (Piecewise with Transformations)
Problem: Let \[ f(x)= \begin{cases} x+2, & x<0 \\ x^2, & x\ge0 \end{cases} \] Apply \( g(x)=-f(x-1)+4 \).
- For \( x-1<0 \) (i.e. \( x<1 \)): \( f(x-1)=(x-1)+2=x+1 \), so \( g(x)=-[x+1]+4=3-x \).
- For \( x-1\ge0 \) (i.e. \( x\ge1 \)): \( f(x-1)=(x-1)^2 \), so \( g(x)=-(x-1)^2+4 \).
Thus, \[ g(x)= \begin{cases} 3-x, & x<1 \\ 4-(x-1)^2, & x\ge1 \end{cases} \]
7) Practice Questions 🎯
7.1 Fundamental – Build Skills (10+)
- Composition: Let \( f(x)=x^2+1 \) and \( g(x)=\sqrt{x-3} \). Find \( (f\circ g)(x) \) and its domain.
- Inverse: For \( f(x)=\frac{3x-5}{2x+4} \), find \( f^{-1}(x) \) with domain and range restrictions.
- Piecewise: Given \[ h(x)=\begin{cases} x+1, & x<2 \\ (x-2)^2, & x\ge2 \end{cases}, \] evaluate \( h(-1) \) and \( h(3) \) and graph roughly.
- Transformations: If \( f(x)=x^3 \), define \( g(x)=-2f(3x+1)+1 \). Describe the transformations step by step.
- Parametric: Show that \( x=\cos t,\ y=\sin t \) traces a unit circle. For which \( t \) does it complete the circle?
- Composition: Let \( f(x)=\ln x \) and \( g(x)=e^x+2 \). Find \( (g\circ f)(x) \) and its domain.
- Inverse: Restrict \( f(x)=x^2-2x+5 \) to \( x\ge1 \) and find its inverse.
- Shift & Scale: Start with \( y=\sqrt{x} \). Write the new function after shifting right 2, reflecting horizontally, scaling vertically by 3, and shifting up 1.
- Piecewise: For \( f(x)=|x| \), define \( g(x)=f(x-3)+2 \). Write \( g(x) \) as a piecewise function and find its domain and range.
- Composition: Let \( f(x)=2^x \) and \( g(x)=x+1 \). Find \( (f\circ g)(x) \) and \( (g\circ f)(x) \); which is larger for large \( x \)?
7.2 Challenging – Push Limits (5+)
- 🔥 Advanced Inverse: If \( f(x)=e^x+e^{-x} \), solve \( f(x)=y \) for \( x \) in terms of \( y \). (Hint: form a quadratic in \( e^x \)).
- 🔥 Piecewise Composition: Given \[ f(x)=\begin{cases} x+2, & x<0 \\ x^2, & x\ge0 \end{cases},\quad g(x)=\begin{cases} -x, & x\le2 \\ 2x-1, & x>2 \end{cases}, \] find \( (f\circ g)(x) \) piecewise.
- 🔥 Parametric: Suppose \( x=t^2 \) and \( y=t^3-t \). Eliminate \( t \) if possible or describe the curve.
- 🔥 Transformations: If \( y=\frac{1}{x} \), apply a transformation that shifts left 1, reflects across the x-axis, and stretches horizontally by a factor of 2. Write the new function.
- 🔥 Composition: Solve \( (f\circ f)(x)=3 \) for \( f(x)=x^2-4 \); that is, solve \( ((x^2-4)^2-4)=3 \) for real \( x \).
8) Summary
- Function composition merges two functions into one, but domain issues must be handled carefully.
- Inverse functions require a one-to-one relationship; sometimes the domain must be restricted.
- Piecewise functions allow different formulas over different intervals.
- Transformations (shifts, scales, reflections) can be combined to produce complex mappings.
- These advanced function concepts enable sophisticated modeling and problem-solving in mathematics.
With these techniques, you can construct and manipulate advanced functions to model real-world scenarios and solve challenging problems. Keep practicing, and enjoy the power of advanced functions! 🌟