Look at the functions in the table below.
Is there a property or feature that is shared by two functions in a row but not the third? If so, we can say that the third function is the odd one out.
Find a reason why each function in the table could be the odd one out in its row.
| \(y=\sin x^2\) | \(y=\ln{x^2}\) | \(y=\tan x (\sec^2 x-1)\) |
| \(y=9x^2-6x+1\) | \(y=\ln{3x}\) | \(y=\sqrt{3x-1}\) |
| \(y=e^{5x}\) | \(y=\dfrac{1}{x^2+4x+4}\) | \(y=e^{x+4}\) |
Here are some reasons why each function is the odd one out. Can you think of others?
\(y=\sin x^2\) \(\dfrac{dy}{dx}=2x\cos x^2\) |
\(y=\ln{x^2}\) \(\dfrac{dy}{dx}=\dfrac{2}{x}\) |
\(y=\tan x (\sec^2 x-1)\) \(\dfrac{dy}{dx}=3\tan^2 x \sec^2 x\) |
If I think about the functions that \(y\) is composed from, I notice that I can use \(u=x^2\) for both \(\sin x^2\) and \(\ln{x^2}\), but not for \(\tan x (\sec^2 x-1).\) For this reason, \(\tan x (\sec^2 x-1)\) is the odd one out. Alternatively, I could say it is the odd one out because \(\sin x^2\) and \(\ln{x^2}\) are both even functions, but \(\tan x (\sec^2 x-1)\) is odd. I could also say that the derivatives of \(\sin x^2\) and \(\ln{x^2}\) can be negative, whereas the derivative of \(\tan x (\sec^2 x-1)\) is never negative.
Continuing to think about the derivatives, I notice that the derivatives of \(\sin x^2\) and \(\tan x (\sec^2 x-1)\) are both zero when \(x=0\), so the graphs of these functions both have stationary points when \(x=0\), whereas the graph of \(\ln{x^2}\) does not. Alternatively, the graphs of \(\sin x^2\) and \(\tan x (\sec^2 x-1)\) have points if inflection while \(\ln{x^2}\) does not.
Finally, I notice that the graphs of \(\ln{x^2}\) and \(\tan^3 x\) have asymptotes, whereas \(\sin x^2\) does not, making \(\sin x^2\) the odd one out. I can also see that the range of \(\sin x^2\) is \([-1,1]\) whereas the range of the other two functions is \(\mathbb{R}.\)
\(y=9x^2-6x+1\) \(\dfrac{dy}{dx}=6(3x-1)\) |
\(y=\ln{3x}\) \(\dfrac{dy}{dx}=\dfrac{1}{x}\) |
\(y=\sqrt{3x-1}\) \(\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-1}}\) |
From the graphs of the functions or by solving \(y=0\), I see that all three graphs meet the \(x\)-axis at \(\tfrac{1}{3}\), but their behaviour at this point is very different. The derivatives of \(\ln{3x}\) and \(9x^2-6x+1\) are both defined when \(x=\tfrac{1}{3}\), but the derivative of \(\sqrt{3x-1}\) is not, making this the odd one out.
As \(9x^2-6x+1=(3x-1)^2\), I can compose both this function and \(\sqrt{3x-1}\) from \(u=3x-1,\) which makes \(\ln{3x}\) the odd one out. The graphs of all three functions involve rescaling a standard function by a factor of \(\tfrac{1}{3}\) parallel to the \(x\)-axis, but \(\ln{3x}=\ln{x}+\ln 3\), so this is the only one which can also be thought of as a translation parallel to the \(y\)-axis.How do the different transformations affect the derivatives of each function?
\(y=e^{5x}\) \(\dfrac{dy}{dx}=5e^{5x}\) |
\(y=\dfrac{1}{x^2+4x+4}\) \(\dfrac{dy}{dx}=\dfrac{-2}{(x+2)^{3}}\) |
\(y=e^{x+4}\) \(\dfrac{dy}{dx}=e^{x+4}\) |
Looking at the gradient functions and graphs, I notice that the gradients of \(y=e^{5x}\) and \(y=e^{x+4}\) are always positive, whereas \(y=\dfrac{1}{x^2+4x+4}\) has negative gradient for \(x>-2.\) If instead I think about composition of functions, I can write \(e^{5x}=(e^x)^{5}\) and \(e^{x+4}=e^4\times e^x\), so these are both compositions of \(e^x\) with another function, making \(\dfrac{1}{x^2+4x+4}\) the odd one out again.
I notice that \(e^{x+4}\) is the only function in the row (and the whole table) whose derivative is equal to itself. Thinking again about composition of functions, if I view \(\dfrac{1}{x^2+4x+4}\) as \((x+2)^{-2}\) then this and \(e^{5x}\) are both powers of a function, whereas \(e^{x+4}\) is not.
Which functions could you have identified as an odd one out from the formula? When did the graph help? How much extra information did finding \(\dfrac{dy}{dx}\) give you?
What features or properties does \(y\) inherit from the functions it’s composed from?
Can you also find an odd one out within each column?
Some of the ideas above can also explain why a function is the odd one out in its column, but there are other reasons too.
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\(y=\sin x^2\) \(\dfrac{dy}{dx}=2x\cos x^2\) |
I could say that \(\sin x^2\) is the odd one out because the other two functions are standard functions scaled by a constant factor — the derivative of \(9x^2-6x+1\) has a scale factor of \(3\) and the derivative of \(e^{5x}\) has a scale factor of \(5.\) How do these factors of \(3\) or \(5\) arise when we find \(\dfrac{dy}{dx}\)? |
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\(y=9x^2-6x+1\) \(\dfrac{dy}{dx}=6(3x-2)\) |
When I look at how the gradients of the functions behave, I notice that the second derivatives of \(e^{5x}\) and \(\sin x^2\) must both depend on \(x\), whereas the second derivative of \(9x^2-6x+1\) is constant. This makes \(9x^2-6x+1\) the odd one out. |
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\(y=e^{5x}\) \(\dfrac{dy}{dx}=5e^{5x}\) |
One reason why \(e^{5x}\) could be the odd one out is that it is the only function in the column whose graph does not have a line of symmetry. |
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\(y=\ln{x^2}\) \(\dfrac{dy}{dx}=\dfrac{2}{x}\) |
I notice that \(y=\ln{x^2}\) is an even function, which could make it the odd one out in the column. It is also the odd one out because the other two functions are translations of standard functions parallel to the axes. |
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\(y=\ln{3x}\) \(\dfrac{dy}{dx}=\dfrac{1}{x}\) |
Thinking again about lines of symmetry makes \(y=\ln{3x}\) the odd one out. Alternatively, it is the only function in this column that only has positive gradient. |
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\(y=\dfrac{1}{x^2+4x+4}\) \(\dfrac{dy}{dx}=\dfrac{-2}{(x+2)^{3}}\) |
The derivatives of \(\ln{x^2}\) and \(\ln{3x}\) both have denominator \(x\), which distinguishes them from \(\dfrac{1}{x^2+4x+4}.\) It is tempting to say that the derivatives of \(\ln{x^2}\) and \(\ln{3x}\) are both odd functions, but \(\ln{3x}\) is only defined for \(x>0.\) |
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\(y=\tan x (\sec^2 x-1)\) \(\dfrac{dy}{dx}=3\tan^2 x \sec^2 x\) |
Either by looking at the derivatives or the graphs, I can see that \(\tan x (\sec^2 x-1)\) is the only function whose graph has stationary points. These are also points of inflection, and neither of the other two functions have points of inflection. In other words, \(\tan x (\sec^2 x-1)\) is the only function whose second derivative changes sign. Note that I don’t have to compute the second derivatives to see this: the graph of \(e^{x+4}\) is always turning to the left and the graph of \(\sqrt{3x-1}\) is always turning to the right, whereas each section of the graph of \(\tan x (\sec^2 x-1)\) changes from turning to the right to turning to the left. This idea is discussed in more detail in Gradients of gradients. |
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\(y=\sqrt{3x-1}\) \(\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-1}}\) |
The behaviour of the gradients of the graphs also give a reason why \(\sqrt{3x-1}\) is the odd one out: it is the only function in the column whose graph gets less steep as \(y\rightarrow \infty.\) |
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\(y=e^{x+4}\) \(\dfrac{dy}{dx}=e^{x+4}\) |
Finally, \(y=e^{x+4}\) is the only function in this column for which the \(x\)-axis is an asymptote. We can see that as \(x\rightarrow -\infty\) the value of the function tends to zero. |
