The Computer Laboratory

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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