Tutorial Week 8

Curve Sketching

There isn’t one single way to do curve sketching, but the following steps present an easy-to-follow method that ensures you’ll have all the information required to sketch:

  1. Domain

  2. Intercepts (x and y intercepts)

  3. Symmetry (odd or even)

  4. Asymptotes (horizontal, vertical, slant/oblique)

  5. Critical points and intervals of increase/decrease (first derivative)

  6. Concavity and inflection points (second derivative)

  7. Sketch

Q1: Sketch \(f(x) = \ln\left(x\right)^{2}\).

../../_images/17.jpg

Derivatives of inverse functions

There’s a helpful formula to find the derivative of an inverse function, that being \((f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))\).

Q2: Given \(f(x) = x^3 + ln(7x + 1) + 1\), find \((f^{-1})'(1)\).

We will use the formula \((f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}\).

We see that first of all, we need \(f^{-1}(x)\). Notice how this function isn’t easily invertible. The strategy now would be to simply try substituting in random integers for x (e.g. -2, -1, 0, 1, 2), and most of the time, it will be one of those since you’re not really expected to actually find the inverse function.

Let \(a = f^{-1}(1)\). This means that \(f(a) = 1\).

We have

\(1 = a^3 + ln(7a + 1) + 1\)

\(0 = a^3 + ln(7a + 1) + 1\)

If we try plugging in -2, -1, 0, 1, and 2 for the value of a, we see that \(a = 0\) is a solution to the equation.

So \(f^{-1}(1) = a = 0\).

Now let’s find \(f'(f^{-1}(1)) = f'(0)\).

\[\begin{split}\begin{aligned} f'(x) &= \frac{d}{dx} x^3 + ln(7x + 1) + 1 \\ &= 3x^2 + \frac{7}{7x + 1} \\ f'(0) &= 3(0)^2 + \frac{7}{7(0) + 1} \\ &= 7 \end{aligned}\end{split}\]

So then

\[\begin{split}\begin{aligned} (f^{-1}(1))' &= \frac{1}{f'(f^{-1}(1))} \\ &= \frac{1}{7} \end{aligned}\end{split}\]