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Get Ready: Module 3 — Applications of Derivatives

Everyone starts somewhere. Before we put the derivative to work, we'll make sure the tool is sharp. This page rebuilds the handful of skills Module 3 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

3 Foundation Skills Worked Examples Self-Paced

Module 3 takes the derivative you learned to compute in Module 2 and puts it to work — finding maximums, minimums, and the shape of a curve. To do that you'll differentiate, then solve \(f'(x)=0\), then reason about which inputs are even allowed. If any of those feel fuzzy, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.

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Skills to build first

Three short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

The derivative rules (from Module 2)

What it is: the toolkit that turns a function into its derivative. The big four are the power rule \(\frac{d}{dx}x^n = n x^{n-1}\), plus the constant-multiple, sum, and (for harder pieces) the product, quotient, and chain rules. Why you need it: every application in this module starts by differentiating. If finding \(f'(x)\) is automatic, you can spend your energy on the interesting part — what the derivative means.

Worked example — differentiate \(f(x)=4x^3-5x^2+7x-2\)
  1. Power rule, term by term. \(\frac{d}{dx}(4x^3)=12x^2\) (bring down the 3, subtract 1 from the exponent, keep the 4).
  2. Next term. \(\frac{d}{dx}(-5x^2)=-10x\).
  3. The linear term. \(\frac{d}{dx}(7x)=7\) (since \(x^1\to 1\cdot x^0=1\)).
  4. The constant. \(\frac{d}{dx}(-2)=0\) — constants have zero slope.

Put it together: \(f'(x)=12x^2-10x+7\). That's the slope of \(f\) at any input \(x\).

Try it 1 — Differentiate \(g(x)=x^3-6x\).
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\(g'(x)=3x^2-6\). (Power rule on \(x^3\) gives \(3x^2\); the derivative of \(-6x\) is \(-6\).)

Try it 2 — Find \(\frac{d}{dx}\big[\,2x^4 - x^2 + 9\,\big]\).
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\(8x^3 - 2x\). (The \(+9\) is a constant, so it contributes \(0\).)

Practice more (free) Khan Academy → Module 2 (Derivatives) →
2 Skill

Solving equations — especially \(f'(x)=0\)

What it is: finding the inputs that make an expression equal to a target value — here, almost always zero. You'll factor, use the zero-product property, or apply the quadratic formula. Why you need it: a critical point is an input where \(f'(x)=0\). Optimization, the first-derivative test, and related rates all come down to setting a derivative equal to zero and solving.

Worked example — solve \(f'(x)=0\) for \(f'(x)=3x^2-12\)
  1. Set it to zero. \(3x^2-12=0\).
  2. Isolate \(x^2\). Add 12, divide by 3: \(x^2=4\).
  3. Take square roots — both signs. \(x=\pm 2\). (Forgetting the negative root loses a critical point!)
  4. State the critical inputs. \(x=-2\) and \(x=2\) — the only places this \(f\) can have a local max or min.

When it factors, use the zero-product property: \(x^2-5x+6=(x-2)(x-3)=0\Rightarrow x=2\) or \(x=3\). When it doesn't, fall back on the quadratic formula \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).

Try it 1 — Solve \(2x^2-8=0\).
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\(2x^2=8\Rightarrow x^2=4\Rightarrow x=\pm 2\). Two solutions — keep both.

Try it 2 — Solve \(x^2-7x+12=0\) by factoring.
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\((x-3)(x-4)=0\), so \(x=3\) or \(x=4\). (Two numbers that multiply to \(12\) and add to \(-7\): \(-3\) and \(-4\).)

Practice more (free) Khan Academy → IXL →
3 Skill

Reasoning about domains & critical values

What it is: figuring out which inputs are allowed (the domain) and which special inputs matter (critical values and interval endpoints). A model often only makes sense on a closed interval — a length can't be negative, a cut can't be bigger than half the sheet. Why you need it: an absolute maximum on a closed interval can hide at an endpoint, not just at a critical point. Knowing the legal domain is what stops you from reporting an impossible answer.

Worked example — the domain for an open-box cut
  1. Read the constraint. A square of side \(x\) is cut from each corner of a \(10\times 10\) sheet.
  2. The cut can't be zero or negative for a real box: \(x>0\).
  3. Two cuts must fit across the sheet: \(2x<10\), so \(x<5\).
  4. Write the domain. \(0\le x\le 5\) (closed, since the endpoints give the degenerate \(V=0\) boxes worth checking).

The candidates for the absolute max are every critical value inside \((0,5)\) plus the endpoints \(x=0\) and \(x=5\). Evaluate the function at all of them and pick the largest.

Try it 1 — A garden's width is \(w\) and its length is \(20-w\). What values of \(w\) make sense?
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Both sides must be positive: \(w>0\) and \(20-w>0\Rightarrow w<20\). So \(0<w<20\). (Endpoints give a zero-area "garden.")

Try it 2 — To find the absolute max of \(f\) on \([0,4]\), which inputs must you test?
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Every critical value (where \(f'=0\)) that lies between \(0\) and \(4\), and the two endpoints \(x=0\) and \(x=4\). Compare \(f\) at all of them.

Practice more (free) Khan Academy → IXL →
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Quick Readiness Check

Six short questions spanning all three skills. Try each one first, then reveal the answer to check yourself.

  1. Differentiate \(f(x)=5x^2-3x+8\).
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    \(f'(x)=10x-3\). (The \(+8\) contributes \(0\).)

  2. Find \(\frac{d}{dx}\big[x^3-12x\big]\).
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    \(3x^2-12\).

  3. Solve \(3x^2-12=0\).
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    \(x^2=4\Rightarrow x=\pm 2\). (Both roots!)

  4. Solve \(x^2-x-6=0\) by factoring.
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    \((x-3)(x+2)=0\), so \(x=3\) or \(x=-2\).

  5. A cut \(x\) is taken from each corner of a \(14\times 14\) sheet. What is the domain of \(x\)?
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    \(0\le x\le 7\) (need \(x>0\) and \(2x<14\)).

  6. To find the absolute max of \(f\) on a closed interval \([a,b]\), what two kinds of inputs must you check?
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    The critical values inside \((a,b)\) and the endpoints \(a\) and \(b\). (Don't forget the endpoints!)

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Tool sharpened. When the skills above feel steady, step into the module — the Visual Lab lets you see the derivative find the largest box and read the shape of a curve.

Module 3 Overview & Visual Lab Jump to the Visual Lab Student Support