OMPT drill — Calculus

Applications of derivatives

Tangent lines, extreme values, and optimisation. The derivative is the tool; the question is really about translating a situation into f'(x) = 0. Say the rule you are using out loud before you differentiate or integrate — it sounds silly and it works.

Tested inOMPT-A OMPT-B OMPT-D OMPT-F

Lesson

This topic is where the derivative stops being an exercise and starts answering questions. Two facts power everything. First, $f' (a)$ is the slope of the tangent line at $x = a$ , so tangent-line questions are two-step affairs: get the slope from the derivative, then build the line through the point. Second, at a peak or a valley of a smooth graph the slope is momentarily zero, so hunting extreme values means solving $f' (x) = 0$ .

The OMPT dresses these up in three costumes: pure tangent-line questions, find-the-extremes questions, and word problems where you must first build the function being optimised. The third costume frightens students most and deserves it least, once the function is written down, the machinery is identical. What the word problems really test is translation: reading "a rectangle has perimeter 20" and producing $A (x) = x (10 - x)$ . The calculus that follows is the same $f' (x) = 0$ you would run anywhere else.

Problem 1Numeric answer

Let $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ . At which value of $x$ does $f$ have its local maximum?

Show the worked solution

Answer: 1

I set $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3) = 0$ , giving critical points $x = 1$ and $x = 3$ . The derivative changes from positive to negative at $x = 1$ (and negative to positive at $x = 3$ ), so the local maximum sits at $x = 1$ . The second derivative confirms it: $f^{''} (1) = 6 - 12 = - 6 < 0$ .

Problem 2Multiple choice

What is the equation of the tangent line to $y = x^{2}$ at $x = 3$ ?

A $y = 6 x - 9$
B $y = 6 x + 9$
C $y = 9 x - 18$
D $y = 6 x$

Show the worked solution

Answer: A — $y = 6 x - 9$

The slope is $y^{'} = 2 x = 6$ at $x = 3$ , and the point of tangency is $(3, 9)$ . Point-slope form gives $y - 9 = 6 (x - 3)$ , so $y = 6 x - 9$ . Checking: at $x = 3$ the line gives $18 - 9 = 9$ , matching the curve.

Problem 3Numeric answer

For $x > 0$ , find the value of $x$ that minimises $S (x) = x + \frac{100}{x}$ .

Show the worked solution

Answer: 10

Differentiating: $S^{'} (x) = 1 - \frac{100}{x ^{2}}$ , which vanishes when $x^{2} = 100$ , so $x = 10$ (the negative root is outside the domain). Since $S^{''} (x) = \frac{200}{x ^{3}} > 0$ there, this is indeed a minimum, with value $S (10) = 20$ .

Problem 4Multiple choice

On which interval is $f (x) = x^{3} - 3 x$ decreasing?

A $(- \infty, - 1)$
B $(- 1, 1)$
C $(1, \infty)$
D $(0, \infty)$

Show the worked solution

Answer: B — $(- 1, 1)$

Decreasing means $f^{'} (x) < 0$ . Here $f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$ , which is negative exactly between the roots, on $(- 1, 1)$ . Outside that interval the parabola $3 x^{2} - 3$ is positive and $f$ climbs.

Problem 5Spot the error

A student classifies the critical points of $f (x) = x^{4} - 4 x^{3}$ . Step 1: $f^{'} (x) = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3)$ , so the critical points are $x = 0$ and $x = 3$ . Step 2: $f^{''} (x) = 12 x^{2} - 24 x$ , so $f^{''} (3) = 36 > 0$ , making $x = 3$ a local minimum. Step 3: $f^{''} (0) = 0$ , so $x = 0$ is also a local minimum. Which step contains the error?

Show the worked solution

Answer: Step 3

Steps 1 and 2 are sound. Step 3 misreads the second-derivative test: $f^{''} (0) = 0$ is inconclusive, not evidence of a minimum. Checking the sign of $f^{'} (x) = 4 x^{2} (x - 3)$ around $x = 0$ : it is negative on both sides (the $x^{2}$ factor never changes sign), so $f$ just keeps decreasing through $x = 0$ — neither a max nor a min.

Problem 6Numeric answer

A ball is thrown upward with height $h (t) = - 4.9 t^{2} + 19.6 t$ metres after $t$ seconds. What maximum height does it reach?

Show the worked solution

Answer: 19.6

The velocity is $h^{'} (t) = - 9.8 t + 19.6$ , which is zero at $t = 2$ seconds — the top of the flight. Substituting back: $h (2) = - 4.9 \cdot 4 + 19.6 \cdot 2 = - 19.6 + 39.2 = 19.6$ metres.