Power functions and first sketches

The course opens with the simplest functions there are, the powers of $x$ , and one question about them: for a given $x$ , which of two powers is bigger? Answering it for very small and very large $x$ is enough to sketch any two-term polynomial, to locate the intersections of power functions, and to see why an exponential eventually beats every power. It is also the language the rest of the course uses for limits at 0 and at infinity.

Start practising Browse all 16 questions

In this topic

Taught: Before the midterm
Questions: 16
Total marks: 35
Suggested time: 61 min
Difficulty: Introductory to Capstone
Show the 3 hardest

Key results

Comparing two powers: $0 < x < 1:\ \ x^{p} < x^{q}, \qquad x > 1:\ \ x^{p} > x^{q}$
For constants $p > q$ .
The dominant term: $a x^{p} + b x^{q} \approx \begin{cases} a x^{p} & \text{for } x \text{ near } 0 \\ b x^{q} & \text{for large } |x| \end{cases}$
For constants $p < q$ and nonzero $a$ , $b$ , with $x > 0$ unless both powers are whole numbers, where $\approx$ means that the ratio of the two sides tends to 1.
Laws of exponents: $x^{a}x^{b} = x^{a + b}, \qquad \frac{x^{a}}{x^{b}} = x^{a - b}, \qquad \bigl(x^{a}\bigr)^{b} = x^{ab}$
For $x > 0$ and any constants $a$ and $b$ .
Sinusoids: $y = A + B\sin\bigl(k(t - c)\bigr):\quad \text{period } \frac{2\pi}{|k|}, \quad A - |B| \le y \le A + |B|$
For constants $A$ , $B$ , $c$ and $k$ , with $B \neq 0$ , $k \neq 0$ and the angle $k(t - c)$ in radians.
Growth in the long run: $(\log x)^{m} < x^{p} < a^{x}$
For any constants $m > 0$ , $p > 0$ and $a > 1$ , once $x$ is large enough.

Read the theoryUBC Math 100 §1.1 Power functionsUBC Math 100 §1.2 First steps in graph sketchingUBC Math 100 §1.5 Familiar functions

Worked example

Standard3 marks · 4 min

The speed of a reaction depends on the concentration $x$ of a chemical according to the function below. As $x$ grows large, $y$ approaches a limiting value. Find the concentration $x > 0$ at which $y$ is exactly half of that limiting value.

y = \frac{6x^{3}}{2 + x^{3}}

Strategy

First find the limiting value by keeping only the dominant terms for large x. Then set y equal to half of it and solve for x.

Working

$\text{Large } x:\quad \frac{6x^{3}}{2 + x^{3}} \approx \frac{6x^{3}}{x^{3}} = 6$
In the denominator, x³ dominates the constant 2. The limiting value is 6.
$\frac{6x^{3}}{2 + x^{3}} = 3 \iff 6x^{3} = 6 + 3x^{3}$
Half of 6 is 3; multiply both sides by the denominator.
$3x^{3} = 6 \iff x^{3} = 2 \iff x = \sqrt[3]{2}$
Collect the x³ terms and take the cube root. x³ is increasing, so x³ = 2 has exactly one real solution, ∛2, and it is positive, as a concentration must be.

Answer

x = \sqrt[3]{2} \approx 1.260

Check

At x = ∛2, x³ = 2 and y = 12/4 = 3, exactly half of the value 6 that y approaches.

Where this goes wrong

Taking the limiting value to be 3 because “6 divided by 2”. The 2 is added to x³, not multiplied; for large x it is swamped, and the fraction tends to 6.

On the paper

Name the term that dominates in each region and why: the lower power near $0$ , the higher one for large $|x|$ . A sketch shows both.

Your turn

Standard2 marks · 4 min

Which description matches the graph of $y = x^{3} - x^{5}$ ?

Answer this question

Practice questions

5 of the 16, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Let $f(x) = 3x^{2} - x^{5}$ . For very large positive $x$ , which term dominates, and what is the sign of $f(x)$ ?
Strategy
For large x the higher power wins, however big the coefficient on the lower power. Decide which term is bigger, then read off its sign.
Working
$x = 10:\quad 3x^{2} = 300, \qquad x^{5} = 100\,000$
Already at x = 10 the fifth power is more than three hundred times the square.
$f(x) = 3x^{2} - x^{5} \approx -x^{5} \quad \text{for large } x$
The graph follows its dominant term.
$f(x) < 0 \quad \text{for large positive } x$
A negative multiple of an odd power is negative when x is positive.
Answer
$f(x) \approx -x^{5} < 0$
The answer is The x to the fifth term dominates, and f(x) is negative for large positive x..
Check
f(x) = x²(3 − x³) is negative once x³ > 3, that is for every x above ∛3 ≈ 1.44.
Where this goes wrong
Letting the coefficient 3 decide. A coefficient only scales a term; for large enough x the higher power always overtakes, whatever the coefficients are.
Try a similar question
Let $g(x) = x^{4} + 6x$ . Close to $x = 0$ , which term dominates, and what does the graph look like there?
Answer this question
Introductory1 mark · 2 min
Let $g(x) = x^{4} + 6x$ . Close to $x = 0$ , which term dominates, and what does the graph look like there?
Strategy
Near zero the roles reverse: the lower power wins, because a number smaller than 1 shrinks each time it is multiplied by itself.
Working
$x = 0.1:\quad x^{4} = 0.0001, \qquad 6x = 0.6$
The linear term is six thousand times the quartic term.
$g(x) = x^{4} + 6x \approx 6x \quad \text{for } x \text{ near } 0$
The graph follows the dominant term, here a straight line of slope 6 through the origin.
Answer
$g(x) \approx 6x \quad \text{near } x = 0$
The answer is The 6x term dominates, and near the origin the graph looks like the line y = 6x..
Check
At x = 0.01, g = 0.06000001, differing from 6x = 0.06 only in the eighth decimal place.
Where this goes wrong
Carrying over the rule for large x. For large x the higher power dominates; close to 0 the lower power does, because there x⁴ is far smaller than x.
Try a similar question
Let $f(x) = 3x^{2} - x^{5}$ . For very large positive $x$ , which term dominates, and what is the sign of $f(x)$ ?
Answer this question
Standard2 marks · 3 min
Find the positive value of $x$ at which the curves below intersect.
$y = x^{4} \qquad \text{and} \qquad y = 5x^{2}$
Strategy
Set the formulas equal and factor out the lowest power of x. The factor that remains, x² − 5, is a quadratic in x whose positive root is the answer.
Working
$x^{4} = 5x^{2} \iff x^{4} - 5x^{2} = 0$
Everything to one side.
$x^{2}\,(x^{2} - 5) = 0$
Factor out x², the lowest power present.
$x = 0 \quad \text{or} \quad x^{2} = 5 \iff x = \pm\sqrt{5}$
Two power functions with positive exponents always meet at the origin; the question wants the positive one.
Answer
$x = \sqrt{5} \approx 2.236$
The answer is the square root of 5, about 2.236.
Check
At x = √5 both curves are at height 25, since x⁴ = 5² and 5x² = 5 · 5.
Where this goes wrong
Answering x = 5 by cancelling x² on both sides and forgetting the equation that remains is x² = 5, not x = 5.
Try a similar question
The curves $y = 2x^{3}$ and $y = 8x$ meet at three points. Find the largest $x$ -coordinate at which they meet.
Answer this question
Challenging3 marks · 5 min
Find the value of $x > 0$ at which the curves below intersect.
$y = 4\sqrt{x} \qquad \text{and} \qquad y = x^{2}$
Strategy
Both curves are power functions, so set them equal and use the laws of exponents to isolate a single power of x. Because x is positive, dividing by a power of x is safe.
Working
$4x^{1/2} = x^{2}$
Write the square root as a fractional power so the exponent laws apply.
$4 = x^{2 - 1/2} = x^{3/2}$
Divide by x^1/2. This would lose x = 0, but the question asks for x > 0.
$x = 4^{2/3} = \bigl(2^{2}\bigr)^{2/3} = 2^{4/3} = \sqrt[3]{16}$
Raise both sides to the reciprocal power 2/3.
$\text{Check: } 4\sqrt{2^{4/3}} = 4 \cdot 2^{2/3} = 2^{2 + 2/3} = 2^{8/3} = \bigl(2^{4/3}\bigr)^{2}$
Both sides agree, so the curves really meet there.
Answer
$x = 4^{2/3} = \sqrt[3]{16} \approx 2.520$
The answer is 4 to the power two-thirds, which is the cube root of 16, about 2.520.
Check
At x = 2.52, 4√x ≈ 6.35 and x² ≈ 6.35, so the two curves do meet there.
Where this goes wrong
Undoing the wrong exponent at the end. Squaring both sides is fine and gives 16x = x⁴, so x³ = 16, but the answer is then written as 16^1/4 or ∛4. The exponent left on x is 3, so x = ∛16.
Try a similar question
Find the positive value of $x$ at which the curves below intersect. $y = x^{4} \qquad \text{and} \qquad y = 5x^{2}$
Answer this question
Capstone5 marks · 9 min
Let $y = \rho x - x^{5}$ , where $\rho$ is a nonzero constant. Which statement about the graph is correct for both signs of $\rho$ ?
Strategy
Treat ρ as a fixed number and run the usual sketch: dominant term near zero, dominant term far away, zeros from factoring. Then ask which of those conclusions change when ρ changes sign.
Working
$\rho x - x^{5} = x\,(\rho - x^{4})$
Factor out the lowest power. The zeros are x = 0 and the solutions of x⁴ = ρ.
$\rho > 0:\ x^{4} = \rho \iff x = \pm \rho^{1/4}; \qquad \rho < 0:\ x^{4} = \rho \text{ has no real solution}$
A fourth power is never negative, so for ρ < 0 the only zero is the origin. For ρ > 0 there are three.
$\text{Near } 0:\quad y \approx \rho x$
The linear term dominates near the origin, so the graph leaves the origin along a line of slope ρ: rising if ρ > 0, falling if ρ < 0.
$\text{Large } |x|:\quad y \approx -x^{5}$
The fifth-power term dominates far away, and it does not involve ρ, so for large positive x the graph falls whatever the sign of ρ.
Answer
$\rho > 0: \text{ zeros } 0, \pm\rho^{1/4},\ \text{rises near } 0;\quad \rho < 0: \text{ zero } 0 \text{ only, falls near } 0;\quad y \approx -x^{5} \text{ far away}$
The answer is Three zeros and a rising start for positive ρ, one zero and a falling start for negative ρ, and a fall like negative x to the fifth for large x in both cases..
Check
ρ = 16 gives zeros 0 and ±2, ρ = −16 gives only 0 since x⁴ + 16 > 0, and y(3) is −195 or −291, negative either way.
Where this goes wrong
Attaching the sign of ρ to the wrong end of the graph. ρ multiplies the low-power term, so it controls the behaviour near the origin and the number of zeros; the far-away behaviour belongs to −x⁵ and never changes.
Try a similar question
Which description matches the graph of $y = x^{3} - x^{5}$ ?
Answer this question

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