Section 04: Advanced Functions
Work individually for 5 minutes, then discuss with the class (5 minutes)
Find the vertex of \(f(x) = 2x^2 - 8x + 3\)
If \(f(x) = 2x + 1\) and \(g(x) = x^2\), find \((f \circ g)(3)\)
Given the transformation \(h(x) = -2f(x - 3) + 4\), describe all transformations applied to \(f(x)\)
Find the inverse of \(f(x) = 3x - 5\)
15 minutes - presentation and discussion
Focus on mock exam preparation and key concepts
Polynomials will extend these concepts to more complex business scenarios!
By the end of this session, you will be able to:
Building on our function knowledge
A polynomial function has the form:
\[P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\]
Key terminology you need to know
Structural Terms:
Degree: highest power of \(x\)
Leading: highest power coefficient
Constant term: \(a_0\)
Standard: descending powers
Examples:
Quick Check
Is \(f(x) = \frac{1}{x} + x^2\) a polynomial? No! The term \(\frac{1}{x} = x^{-1}\) has a negative exponent.
These two values determine the big picture
The degree tells us:
The leading coefficient determines:
What happens as \(x \to \pm\infty\)?
End behavior depends on:
Even Degree:
Odd Degree:
The four fundamental patterns
2 minutes individual, 3 minutes pairs, 2 minutes class discussion
Without graphing, describe the end behavior:
\(P(x) = -2x^5 + 3x^3 - x + 7\)
\(Q(x) = 4x^6 - x^4 + 2x^2 - 1\)
\(R(x) = -\frac{1}{2}x^4 + 5x^2 + 3\)
Where the polynomial crosses or touches the x-axis
A zero of \(P(x)\) is a value \(c\) where \(P(c) = 0\). To find zeros:
Fundamental Theorem of Algebra
A polynomial of degree \(n\) has exactly \(n\) zeros (counting multiplicities and complex zeros).
A tool for finding zeros: Intermediate Value Theorem (IVT)
If \(P(x)\) is continuous on \([a; b]\) and \(P(a) \cdot P(b) < 0\), then there exists at least one \(c\) in \((a; b)\) where \(P(c) = 0\)
Every polynomial is continuous: its graph has no gaps or jumps.
In simple terms:
IVT guarantees at least one zero exists but doesn’t tell us exactly where or how many!
Locating zeros systematically
Example: Show that \(P(x) = x^3 - 2x - 5\) has a zero in \([2; 3]\)
Solution:
Business Application: Use IVT to prove break-even points exist when you know profit is negative at low production and positive at high production.
Which “easy” numbers make our polynomial equal zero?
Consider any polynomial with integer coefficients, like:
\[P(x) = 2x^3 - 5x^2 + x + 2\]
Question: What values of \(x\) make \(P(x) = 0\)?
Some zeros might be “messy” (like \(\sqrt{2}\) or complex numbers), but some might be “easy” rational numbers (fractions).
Why This Matters?
Instead of guessing randomly, use a systematic approach
The Rule: If our polynomial has integer coefficients, then any rational zero \(\frac{p}{q}\) (in lowest terms) must follow a pattern.
Why p and q?
Let’s use our example: \(P(x) = 6x^3 - 11x^2 + 6x - 1\)
Constant term (the number without \(x\)): -1
Leading coefficient (number in front of highest power): 6
See the Difference?
Mix and match the p and q values
All possible \(\frac{p}{q}\) combinations:
From \(p \in \{\pm 1\}\) and \(q \in \{\pm 1, \pm 2, \pm 3, \pm 6\}\):
\[\frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 1}{\pm 3}, \frac{\pm 1}{\pm 6}\]
Smart Testing Order
Start with simple fractions: \(\pm 1\), then try: \(\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}\)
Substitute each \(\frac{p}{q}\) candidate into the polynomial
\[P(1) = 6(1)^3 - 11(1)^2 + 6(1) - 1 = 0 \checkmark\]
\[P(\frac{1}{2}) = 6(\frac{1}{2})^3 - 11(\frac{1}{2})^2 + 6(\frac{1}{2}) - 1 = 0 \checkmark\]
\[P(\frac{1}{3}) = 6(\frac{1}{3})^3 - 11(\frac{1}{3})^2 + 6(\frac{1}{3}) - 1 = 0 \checkmark\]
Result: The rational zeros are \(x = 1\), \(x = \frac{1}{2}\), and \(x = \frac{1}{3}\)
The theorem doesn’t guarantee these will be zeros - it just tells us which ones are worth testing!
Found one zero? Divide to get the rest
If \(P(c) = 0\), then \((x - c)\) is a factor of \(P(x)\):
\[P(x) = (x - c) \cdot Q(x)\]
Typical exam task: profit \(P(x)\) is a cubic. Test \(x = 1, 2, 3, \dots\) until \(P(c) = 0\), divide by \((x - c)\), then solve the quadratic for the other break-even points.
Example: \(P(x) = x^3 - 6x^2 + 3x + 10\), and \(P(2) = 8 - 24 + 6 + 10 = 0\)
\[\begin{array}{l} (x^3 - 6x^2 + 3x + 10) : (x - 2) = x^2 - 4x - 5 \\ \underline{-(x^3 - 2x^2)} \\ \phantom{(x^3}\;\; -4x^2 + 3x \\ \phantom{(x^3}\;\; \underline{-(-4x^2 + 8x)} \\ \phantom{(x^3 - 4x^2}\;\; -5x + 10 \\ \phantom{(x^3 - 4x^2}\;\; \underline{-(-5x + 10)} \\ \phantom{(x^3 - 4x^2 - 5x + 1}\;\; 0 \end{array}\]
Repeat until the remainder has a lower degree than the divisor:
For a cubic divided by \((x - c)\) this takes exactly three rounds and the remainder must be \(0\) when \(c\) is a zero.
Always multiply back - the remainder must be 0
\[\begin{aligned}(x - 2)(x^2 - 4x - 5) &= x^3 - 4x^2 - 5x - 2x^2 + 8x + 10 \\ &= x^3 - 6x^2 + 3x + 10 \checkmark\end{aligned}\]
Factor the quadratic: \(x^2 - 4x - 5 = (x - 5)(x + 1)\)
Complete factorization: \(P(x) = (x - 2)(x - 5)(x + 1)\)
Zeros: \(x = -1\), \(x = 2\), \(x = 5\)
A remainder \(\neq 0\) means \(c\) was not a zero - or an arithmetic slip. Check \(P(c)\) again.
Work alone for 5 minutes, then compare with your neighbor
A company’s profit (in thousands EUR) at production level \(x\) (thousands of units) is
\[P(x) = -x^3 + 8x^2 - 19x + 12\]
The company breaks even at \(x = 1\). Find the other break-even points.
How many times a zero appears
If \((x - c)^m\) is a factor of \(P(x)\), then \(c\) has multiplicity \(m\)
Behavior at zeros:
Reading the story from the factors
Given: \(P(x) = -2(x + 3)(x - 1)^2(x - 4)\)
What can we determine?
A systematic approach
Step-by-step process:
Between zeros, the polynomial doesn’t cross the x-axis. Use test points to determine if the graph is above or below the x-axis in each interval.
Work alone for 5 minutes, then discuss for 5 minutes
Work alone for 5 minutes, then discuss for 5 minutes
Where do we see polynomials in business?
Let’s do this with an example!
When products interact
TechCo produces three related products with profit function:
\[P(x) = -x^3 + 12x^2 - 35x + 24\]
where \(x\) is production level (thousands of units).
Business Question: At what levels does the company break even?
Mathematical Task: Solve \(P(x) = 0\) to find where profit equals zero!
Solving \(P(x) = 0\) to find where profit equals zero
Step 1: Factor out any common factors
\[-(x^3 - 12x^2 + 35x - 24) = 0\]
Step 2: Use the Rational Root Theorem to find possible roots
Solving \(P(x) = 0\) to find where profit equals zero
Step 3: Test \(x = 1\):
\[P(1) = -(1)^3 + 12(1)^2 - 35(1) + 24 = -1 + 12 - 35 + 24 = 0\]
Step 4: Divide by \((x - 1)\) (long division): \(P(x) = -(x - 1)(x^2 - 11x + 24)\)
Step 5: Factor the quadratic: \(x^2 - 11x + 24 = (x - 3)(x - 8)\)
Final form: \(P(x) = -(x - 1)(x - 3)(x - 8)\)
Break-even points: \(x = 1, 3, 8\) thousand units
Analysis: Break-even at 1,000, 3,000, and 8,000 units
Complex cost structures
A manufacturing plant has cost function:
\[C(x) = 0.1x^4 - 2x^3 + 12x^2 + 50\]
Polynomial cost functions capture the reality that unit costs often decrease initially (economies of scale) but may increase at very high production levels (capacity constraints).
Competition over time
Three companies compete with market shares modeled by:
where \(t\) is years from product launch, \(0 \leq t \leq 4\).
We can analyze the market dynamics by analyzing the properties.
End Behavior Analysis: Their competitor has profit function \(C(x) = -3x^5 + 2x^4 - 7x^2 + x - 9\). Describe the long-term behavior as production increases. What does this tell management?
Product Line Analysis: A subsidiary’s profit is modeled by \(S(x) = 2(x + 2)^2(x - 3)(x - 5)^3\). Find all break-even points and describe how the company enters/exits profitability at each point.
New Product Launch: TechCo’s profit (in thousands EUR) for a new product after \(x\) months is: \(P(x) = -x^3 + 9x^2 - 15x - 25\). What is the initial financial position at launch and how is the profit at months 5 and 7?
Can you find the errors? Work with your neighbor
Time allocation: 5 minutes to find errors, 5 minutes to discuss
Student work:
“The polynomial \(P(x) = 3x^4 - 2x^2 + 1\) has degree 2 because there are two terms with \(x\)”
“If \((x - 2)^4\) is a factor, the graph crosses the x-axis at \(x = 2\)”
“A degree 5 polynomial always has 5 real zeros”
A company builds barbecue smokers (grills that cook food slowly in smoke) and sells them to garden centres. Its monthly total costs are
\[C(x) = 0.5x^3 - 4x^2 + 20x + 72\]
with \(C\) in MU and \(x\) in QU per month (1 QU = 100 smokers, 1 MU = 1,000 EUR). The garden centres pay 300 EUR per smoker.
(for verification: \(P(x) = -0.5x^3 + 4x^2 + 10x - 72\))
A competitor forces the company to cut its price to 260 EUR per smoker. The costs stay the same.
Show that the new profit function is \(P_2(x) = -0.5(x - 6)^2(x + 4)\). Give its zeros with their multiplicities and describe the graph at \(x = 6\). The sales manager says: “If we just sell enough, we still make a profit.” Assess this claim. [5 pts.]
Compute the average cost per smoker (total costs divided by the number of smokers) at 600 smokers. Explain with this result why the profit at 600 smokers is exactly 0 EUR. [3 pts.]
Today’s essential concepts
5 minutes - Individual work
Given the polynomial \(P(x) = -2x^3 + 6x^2 + 8x\):
Factor completely and find all zeros with their multiplicities
Determine the end behavior
Describe the graph’s behavior at each zero
If this represents a company’s profit (in thousands EUR) where \(x\) is production in thousands of units, at what production levels does the company break even?
Session 04-02: Power, Root & Rational Functions
Building on polynomial foundations
Complete Tasks 04-01!
Session 04-01 - Polynomial Functions | Dr. Nikolai Heinrichs & Dr. Tobias Vlćek | Home