$f(x),$ 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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- Evaluate
$\frac{d}{dx}{(x^2-3)^3}.$ - Find the roots of the polynomial
$x^2-7x+12.$ - Find the inflexion point of the curve
$y=x^3-3x^2+5x-1.$
- Evaluate
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What does
$f$being symmetrical about the$y$-axis imply about its degree$n?$ -
Does the fact that
$f$has a point of inflexion restrict the values that$n$can take? -
How can one express a polynomial in terms of its roots?
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If you arbitrarily choose one root of the polynomial, what conditions does that set on other roots?
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Once you have arbitrarily selected a pair of roots, how can you select the other pair to satisfy the inflexion point condition?
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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.