The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. # Practice Paper 3 Question 16

Give an example (i.e. equation) of a non-constant polynomial function of the smallest degree $$f(x),$$ such that all of the following hold: (a) it has at least one inflexion point, (b) all inflexion points have $$y$$-coordinate equal to $$0,$$ (c) all its roots are real, and (d) it is symmetric with respect to the $$y$$-axis.

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## Warm-up Questions

1. Evaluate $$\frac{d}{dx}{(x^2-3)^3}.$$
2. Find the roots of the polynomial $$x^2-7x+12.$$
3. Find the inflexion point of the curve $$y=x^3-3x^2+5x-1.$$

## Hints

• Hint 1
What does $$f$$ being symmetrical about the $$y$$-axis imply about its degree $$n?$$
• Hint 2
Does the fact that $$f$$ has a point of inflexion restrict the values that $$n$$ can take?
• Hint 3
How can one express a polynomial in terms of its roots?
• Hint 4
If you arbitrarily choose one root of the polynomial, what conditions does that set on other roots?
• Hint 5
Once you have arbitrarily selected a pair of roots, how can you select the other pair to satisfy the inflexion point condition?

## Solution

Since $$f$$ is symmetrical about the $$y$$-axis, it must be of even order $$n.$$ At an inflexion point, the second derivative changes sign. For $$n \leq 2,\,$$ $$f''(x)=0$$ and so we deduce $$n > 2.$$ Hence, the smallest possible degree is $$n=4.$$ If the $$y$$-coordinate equals $$0$$ at all the inflexion points, then every inflexion point is a root. If all roots are real, then the polynomial may be expressed as a product of real monomials, i.e. $$f(x)=(x-a)(x-b)\cdots.$$ The symmetry of the function also means that the roots must be symmetrical, so $$f(x)=(x-a)(x+a)(x-b)(x+b)$$$$=(x^2-a^2)(x^2-b^2).$$

We can arbitrarily choose a pair of symmetric roots, so let $$b=1.$$ We then find the second derivative of $$f,$$ which is $$f''(x)=12x^2-2a^2-2.$$ The roots of this derivative, $$\pm\sqrt{(a^2+1)/6},$$ are points of inflexion. In order to satisfy the condition that all inflexion points are roots of $$f,$$ we equate these to $$\pm a$$ and solve to get $$a=\frac{1}{\sqrt5}.$$ Therefore $$f(x)=(x^2-1)(x^2-1/5)$$ in this case.

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