Power functions and first sketches, question 15

Topic: Power functions and first sketches
Difficulty: Capstone
Suggested time: 9 min
Marks: 5
Theory: UBC Math 100 §1.2Key results

Let $y = \rho x - x^{5}$ , where $\rho$ is a nonzero constant. Which statement about the graph is correct for both signs of $\rho$ ?

If ρ > 0 the only real zero is x = 0, and if ρ < 0 there are three, at x = 0 and x = ±(−ρ)^1/4. In both cases the graph looks like −x⁵ near the origin, flat and falling through it, and for large positive x it follows the line y = ρx.
If ρ > 0 there are three real zeros, at x = 0 and x = ±ρ^1/4, and near the origin the graph rises like the line y = ρx. If ρ < 0 the only real zero is x = 0 and near the origin the graph falls like y = ρx. In both cases the graph falls like −x⁵ for large positive x.
For every nonzero ρ there are three real zeros, at x = 0 and the two real solutions of x⁴ = ρ. Near the origin the graph rises like y = ρx when ρ > 0 and falls like y = ρx when ρ < 0, and it falls like −x⁵ for large positive x.
If ρ > 0 there are three real zeros, at x = 0 and x = ±ρ^1/4, and if ρ < 0 the only real zero is x = 0. In both cases the graph looks like −x⁵ near the origin, and the sign of ρ decides whether it rises or falls for large positive x, like y = ρx.

How sure are you?

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Theory UBC Math 100 §1.2Key results

Worked solution

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}

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}

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Power functions and first sketches practice question 15