Advanced Higher Maths
Functions and Graphs

In this simple first example, the numerator is irreducible. So we just have to consider values of \(\raise 0.2pt{x}\) that result in the denominator equalling zero.

$$ x^2-x-6 = (x+2)(x-3) $$

So there are two vertical asymptotes: \(\raise 0.2pt{x\!=\!-2}\) and \(\raise 0.2pt{x\!=\!3\small.}\)

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\(\raise 0.2pt{x\!=\!0}\) and \(\raise 0.2pt{x\!=\!1}\) give rise to a zero denominator. However, \(\raise 0.2pt{\text{sin}\,0=0}\) so \(\raise 0.2pt{x\!=\!0}\) is not a vertical asymptote. If a certain value of \(\raise 0.2pt{x}\) results in both the numerator and denominator being zero, it is not an asymptote.

So there is only one vertical asymptote: \(\raise 0.2pt{x\!=\!1\small.}\)

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There are vertical asymptotes when \(\raise 0.2pt{x^2\!-\!2\!=\!0\small,}\) so when \(\raise 0.2pt{x\!=\!-\sqrt{2}}\) and \(\raise 0.2pt{x\!=\!\sqrt{2}\small.}\)

The degree of the numerator equals the degree of the denominator so there will also be a horizontal asymptote.

Rather than taking the time to set out a polynomial division, the equation of the horizontal asymptote can be seen by dividing \(\raise 0.2pt{f(x)}\) top and bottom by \(\raise 0.2pt{x^2\small.}\)

$$ f(x)=\small\frac{3x^2+2}{x^2-2}\normalsize=\small\frac{3+\large\frac{2}{x^2}\normalsize}{1-\large\frac{2}{x^2}\normalsize}\normalsize $$

As \(\raise 0.2pt{x\rightarrow\infty\small,}\) \(\raise 0.2pt{f(x)\rightarrow\large\frac{3+0}{1-0}\normalsize=3\small.}\)

So there is a horizontal asymptote \(\raise 0.2pt{y\!=\!3.}\)

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\(\raise 0.2pt{f(x)}\) factorises as follows:

$$ f(x)=\small\frac{x(x+1)(x-1)}{(x+2)(x-4)} $$

There are vertical asymptotes when \(\raise 0.2pt{x\!+\!2\!=\!0}\) and when \(\raise 0.2pt{x\!-\!4\!=\!0\small,}\) so at \(\raise 0.2pt{x\!=\!-2}\) and \(\raise 0.2pt{x\!=\!4\small.}\)

The degree of the numerator is one higher than the degree of the denominator, so there will also be an oblique asymptote. To identify its equation, we use polynomial long division:

So the oblique asymptote has equation \(\raise 0.2pt{y\!=\!x\!+\!2.}\)

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To test whether \(f\) is odd, even or neither, we need to obtain an expression for \(\raise 0.2pt{f(-x)\small.}\)

$$ \begin{eqnarray} f(-x) &=& (-x)^2+n \\[6pt] &=& x^2+n \\[6pt] &=& f(x) \end{eqnarray} $$

The condition \(\raise 0.3pt{f(-x)=f(x)\ \forall x}\) defines an even function, so \(f\) is even.

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To test whether \(f\) is odd, even or neither, we need to obtain an expression for \(\raise 0.2pt{f(-x)\small.}\)

$$ \begin{eqnarray} f(-x) &=& (-x)^3\,\text{cos}\,(-x) \\[6pt] &=& -\!x^3\,\text{cos}\,x \\[6pt] &=& -\!f(x) \end{eqnarray} $$

If you are struggling to see why we were able to equate \(\raise 0.3pt{\text{cos}(-x)}\) and \(\raise 0.2pt{\text{cos}\,x}\) above, consider a quadrant diagram for acute \(\raise 0.2pt{x\small.}\) Then \(\raise 0.3pt{x}\) is in the first quadrant, which is positive, and \(\raise 0.3pt{-x}\) is in the fourth quadrant, which is also positive. A similar argument applies for obtuse or reflex \(\raise 0.3pt{x\small.}\) Cosine is an even function.

The condition \(\raise 0.3pt{f(-x)=-f(x)\ \forall x}\) defines an odd function, so \(f\) is odd.

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To test whether \(f\) is odd, even or neither, we need to obtain an expression for \(\raise 0.2pt{f(-x)\small.}\)

$$ \begin{eqnarray} f(-x) &=& e^{2(-x)} \\[6pt] &=& e^{-2x}\\[6pt] &=& \small\frac{1}{e^{2x}}\\[6pt] &=& \small\frac{1}{f(x)} \end{eqnarray} $$

Because \(\raise 0.3pt{f(-x)\neq f(x)}\) and \(\raise 0.3pt{f(-x)\neq -f(x)\small,}\) \(f\) is neither even nor odd.

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To find any points of inflection, we need to consider the second derivative, so let's differentiate twice, using the quotient rule.

$$ \begin{matrix} u=3x^2+2 \:&\: v=x^2-2 \\[5pt] u'=6x \:&\: v'=2x \\ \end{matrix} $$

$$ \begin{eqnarray} f'(x) &=& \frac{u'\,v-u\,v'}{v^2} \\[8pt] &=& \frac{6x(x^2-2)-2x(3x^2+2)}{\left(x^2-2\right)^2} \\[8pt] &=& \frac{6x^3-12x-6x^3-4x}{\left(x^2-2\right)^2} \\[8pt] &=& \frac{-16x}{\left(x^2-2\right)^2} \\[8pt] \end{eqnarray} $$

Now for the second derivative. For \(v'\) we should use the chain rule as this enables us to leave terms as multiples of \(\raise 0.2pt{x^2-2\small,}\) which really helps with the simplification.

$$ \begin{matrix} u=-16x \:&\: v=(x^2-2)^2 \\[5pt] u'=-16 \:&\: v'=2(x^2-2)(2x) \\[4pt] \:&\: =4x(x^2-2) \\ \end{matrix} $$

$$ \begin{eqnarray} f''(x) &=& \frac{u'\,v-u\,v'}{v^2} \\[8pt] &=& \frac{-16(x^2-2)^2+16x(4x(x^2-2))}{\left(x^2-2\right)^4} \\[8pt] &=& \frac{(x^2-2)\left(-16(x^2-2)+16x(4x)\right)}{\left(x^2-2\right)^4} \\[8pt] &=& \frac{-16(x^2-2)+16x(4x)}{\left(x^2-2\right)^3} \\[8pt] &=& \frac{-16x^2+32+64x^2}{\left(x^2-2\right)^3} \\[8pt] &=& \frac{48x^2+32}{\left(x^2-2\right)^3} \\[8pt] \end{eqnarray} $$

A necessary (but not sufficient) condition for a point of inflection is that \(f''(x)=0\) or is undefined, which it never is. Also, \(48x^2+32\) can never equal zero as \(x^2\geqslant 0\ \forall x\small.\)

So the graph of \(y=f(x)\) has no points of inflection.

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To find stationary points, we solve \(\large\frac{dy}{dx}\normalsize\!=\!0\small,\) just like at Higher.

$$ \begin{eqnarray} \small\frac{dy}{dx}\normalsize &=& 6x^2-24x-30 \\[6pt] &=& 6(x^2-4x-5) \\[6pt] &=& 6(x+1)(x-5) \\[6pt] \end{eqnarray} $$

So there are stationary points when \(\raise 0.1pt{x\!=\!-1}\) and \(\raise 0.1pt{x\!=\!5\small.}\)

When \(\raise 0.2pt{x\!=\!-1\small,}\) \(\raise 0.2pt{y\!=\!-2\!-\!12\!+\!30\!+\!9=25\small.}\)

When \(\raise 0.2pt{x\!=\!5\small,}\) \(\raise 0.2pt{y\!=\!250\!-\!300\!-\!150\!+\!9=-191\small.}\)

So the stationary points are \((-1,\,25)\) and \((5,\,-191)\small.\)

To determine their natures, at Advanced Higher, we should use the second derivative:

$$ \small\frac{d^{2}y}{dx^2}\normalsize = 12x-24 $$

When \(\raise 0.2pt{x\!=\!-1\small,}\) \(\large\frac{d^{2}y}{dx^2}\normalsize = -12-24\lt 0 \) so \((-1,\,25)\) is a maximum turning point.

When \(\raise 0.2pt{x\!=\!5\small,}\) \(\large\frac{d^{2}y}{dx^2}\normalsize = 60-24\gt 0 \) so \((5,\,-191)\) is a minimum turning point.

This question also asked about any points of inflection, and luckily we've done most of the work for that already. Points of inflection occur when the second derivative is zero or undefined. As \(12x-24\) is defined for all real \(\raise 0.2pt{x\small,}\) we only need to consider \(12x-24=0\small,\) which occurs when \(\raise 0.2pt{x\!=\!2\small.}\)

As \(\raise 0.2pt{x\!=\!2}\) was not a solution of \(\large\frac{dy}{dx}\normalsize\!=\!0\small,\) the point of inflection is non-horizontal.

When \(\raise 0.2pt{x\!=\!2\small,}\) \(\raise 0.2pt{y\!=\!16\!-\!48\!-\!60\!+\!9=-83\small.}\)

So \((2,\,-83)\) is a non-horizontal point of inflection.

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First we differentiate twice, and manipulate the expression for \(\large\frac{d^{2}y}{dx^2}\) into the 'friendliest' possible form.

$$ \begin{eqnarray} y &=& \text{sin}\,x + \text{tan}\,x\\[6pt] \small\frac{dy}{dx}\normalsize &=& \text{cos}\,x+\text{sec}^{2}\,x \\[6pt] \small\frac{d^{2}y}{dx^2}\normalsize &=& -\!\text{sin}\,x+(2\,\text{sec}\,x)(\text{sec}\,x\,\text{tan}\,x) \\[6pt] &=& -\!\text{sin}\,x+2\,\text{sec}^{2}\,x\,\text{tan}\,x \\[6pt] &=& -\!\text{sin}\,x+2\small\left(\frac{1}{\text{cos}^{2}\,x}\right)\normalsize\small\left(\frac{\text{sin}\,x}{\text{cos}\,x}\right)\normalsize \\[6pt] &=& -\!\text{sin}\,x+\small\frac{2\,\text{sin}\,x}{\text{cos}^{3}\,x} \\[14pt] \end{eqnarray} $$

Now we solve \(\large\frac{d^{2}y}{dx^2}\normalsize =0\small,\) as follows:

$$ \begin{gather} -\text{sin}\,x+\small\frac{2\,\text{sin}\,x}{\text{cos}^{3}\,x}\normalsize = 0 \\[6pt] -\text{sin}\,x\,\text{cos}^{3}\,x+2\,\text{sin}\,x = 0 \\[6pt] (\text{sin}\,x)(-\text{cos}^{3}\,x+2) = 0 \\[6pt] \end{gather} $$

$$ \begin{eqnarray} \text{sin}\,x=0\:\:\ &\small\textsf{or}\normalsize& \:\,\text{cos}^{3}\,x=2 \\[6pt] x=0\:\:\ &\small\textsf{or}\normalsize& \:\,\text{cos}\,x\approx 1.26 \\[6pt] \:\: &\,& \:\, \small\textsf{No solutions}\normalsize\\[6pt] \end{eqnarray} $$

If there is an inflexion point when \(x=0\small,\) the second derivative will change sign when \(x=0\small.\) A table of signs will let us see whether this is the case.

$$ \begin{array} {|c|c|} \hline x & \rightarrow & 0 & \rightarrow \\ \hline \large\frac{d^{2}y}{dx^2}\normalsize & - & 0 & + \\ \hline \end{array} $$

So the point with \(x=0\) is indeed a point of inflexion, and all that remains is to find its \(y\)-coordinate and write our conclusion.

$$ \begin{eqnarray} y &=& \text{sin}\,x + \text{tan}\,x\\[6pt] &=& \text{sin}\,0 + \text{tan}\,0 \\[6pt] &=& 0+0 \\[6pt] &=& 0 \\[6pt] \end{eqnarray} $$

So the point of inflexion is \((0,\,0)\small.\)

Side note: The spellings 'inflection' and 'inflexion' are both correct. Both have been used in exams.

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(a)  The function \(f\) is even. You could give either of the following justifications.

Either note that the graph of \(y=f(x)\) is symmetrical about the y-axis, or demonstrate that \(f(–x)=f(x)\small.\)

(b)  The section of the graph with negative \(y\)-values reflects in the \(x\)-axis, thus:

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(a)  To test whether \(f\) is odd, even or neither, we need to obtain an expression for \(\raise 0.2pt{f(-x)\small.}\)

$$ \begin{eqnarray} f(-x) &=& (-x)^3-(-x) \\[6pt] &=& -\!x^3+x \\[6pt] &=& -\!(x^3-x) \\[6pt] &=& -\!f(x) \end{eqnarray} $$

The function \(f\) is therefore odd.

(b)  To find any points of inflection, we need to consider the second derivative, so let's differentiate twice:

$$ f'(x)=3x^2-1 $$

$$ f''(x)=6x $$

A necessary, but not sufficient, condition for a point of inflection is that \(f''(x)=0\) or is undefined.

Clearly \(f''(x)=0\) when \(x=0\small,\) so we test further to see if this is a point of inflection.

The easiest way of doing this is by checking the signs of \(f''(x)\) just to the left and right of \(x=0\small.\)

When \(x\lt 0\small,\normalsize\:6x\lt 0\) and when \(x\gt 0\small,\normalsize\:6x\gt 0\small.\)

So our second derivative changes sign (a change in concavity) when \(x=0\small.\)

We can therefore conclude that there is a point of inflection at \((0,0)\small.\)

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