Question 11
Let $f(x)=\tan^{-1}\left(\dfrac{x}{2}\right)$ for suitable values of $x$ where $f(x)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.
Determine $f(2)$.
[1 mark] b)Determine $f'(2)$.
[2 marks] c)Use the results from Questions 11a) and 11b) to determine the equation of the tangent to the graph of $y=f(x)$ at $x=2$.
[2 marks]A point, a gradient, then the tangent
The curve is drawn first. Then each part of the question adds one thing to the picture, in the order you need them.
The two axes use different scales, so the tangent looks steeper than $\frac14$. Count the units instead: across $4$ and up $1$.
Move the point and watch the tangent and its gradient
The top graph is the curve with the orange tangent at your point. The bottom graph is the gradient function, so the height of its black point is the gradient of the tangent above it.
Top: $y=f(x)$. Bottom, in blue: $y=f'(x)$. The two graphs use different vertical scales.
This is the question’s point, $\left(2,\frac{\pi}{4}\right)$. The tangent has gradient $\frac14$ and meets the $y$-axis at $\frac{\pi}{4}-\frac12\approx0.29$.
At $x=-2$ the gradient is also $\frac14$, because $f'(x)$ only depends on $x^2$. This tangent is parallel to the question’s one.
At $x=0$ the gradient is $\frac12$, the largest it gets, because $4+x^2$ is smallest there. The tangent is $y=\frac12x$.
The gradient function $\frac{2}{4+x^2}$ is positive for every $x$, so every tangent slopes upwards. Away from $x=0$ the gradient shrinks and the curve flattens towards its asymptotes.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide.
Substitute $x=2$. The inside of the inverse tangent becomes $1$.
$$\begin{aligned}f(2)&=\tan^{-1}\left(\frac{2}{2}\right)\\&=\tan^{-1}(1)\end{aligned}$$
You want the angle whose tangent is $1$. Both $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ have that tangent, but the question says $f(x)$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so only $\frac{\pi}{4}$ fits.
Marker: correctly determines the required value
The formula book has the derivative of the inverse tangent in exactly this form.
$$\begin{aligned}&\frac{d}{dx}\left(\tan^{-1}\left(\frac{x}{a}\right)\right)\\&\quad=\frac{a}{a^2+x^2}\end{aligned}$$
Here $a=2$, so the $2$ goes on top as well as inside the square.
$$f'(x)=\frac{2}{2^2+x^2}=\frac{2}{4+x^2}$$
You get the same thing from the chain rule, with $u=\frac{x}{2}$ and $\frac{d}{du}\tan^{-1}(u)=\frac{1}{1+u^2}$.
$$\begin{aligned}f'(x)&=\frac{1}{1+\left(\frac{x}{2}\right)^2}\times\frac12\\&=\frac{1}{2+\frac{x^2}{2}}\\&=\frac{2}{4+x^2}\end{aligned}$$
Marker: correctly determines the gradient function
Substitute $x=2$ into the gradient function.
$$f'(2)=\frac{2}{4+2^2}=\frac{2}{8}$$
The QCAA also accept $\frac28$ or $0.25$. If your gradient function was wrong, you can still earn this mark by follow-through.
Marker: determines gradient of the tangent
The tangent is a straight line, $y=mx+c$. Part a) gives the point $\left(2,\frac{\pi}{4}\right)$ and part b) gives $m=\frac14$. Substitute both to find $c$.
$$\frac{\pi}{4}=\frac14(2)+c$$
$$c=\frac{\pi}{4}-\frac12$$
The QCAA also accept $\frac{\pi-2}{4}$. If you use $y-y_1=m(x-x_1)$ instead, you never write $c$ on its own, and this mark is implied by your working.
Marker: determines $y$-intercept of the tangent
Put $m$ and $c$ back into the line.
The point-gradient form gets there too.
$$y-\frac{\pi}{4}=\frac14(x-2)$$
Any equivalent equation is accepted, such as $x-4y+\pi=2$. Keep $\pi$ exact. Paper 1 has no calculator, and the marking guide’s answer is exact.
Marker: determines equation of the tangent
Each phrase of the question gave you something
Each part takes one routine move, and the question tells you to use parts a) and b) for part c). The derivative comes straight off the formula book, and the tangent is a straight line through a point you already have.
If I had to guess where you might lose marks, it would be the top of the derivative. The formula book puts $a$ in the numerator, so the derivative is $\frac{2}{4+x^2}$. Writing $\frac{1}{4+x^2}$ gives a gradient of $\frac18$ instead of $\frac14$. Also give $f(2)$ in radians. $45^\circ$ is the same angle in degrees, and degrees can’t sit in the equation of a line next to $\frac14x$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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