Calculus I · Maxima, Minima, and the Shape of a Function

Where this is heading

In Chapter 9, gradient descent chases the condition this chapter is built on — it steps downhill precisely to drive the slope $f'$ toward zero, the critical-point equation you’ve been solving by hand. And the convexity you proved here is exactly the guarantee that its descent can’t get lost. The two chapters are one idea from two sides: this one finds minima with algebra, the next one finds them by iteration when algebra can’t.

And the whole story reopens, larger, in Book 3. There, $x$ becomes a vector, the single slope $f'$ becomes the gradient, “$f' = 0$” becomes “the gradient vanishes,” the second derivative becomes a matrix of curvatures, and convexity becomes a statement about that matrix. The shape of a function in many dimensions is where optimization, and machine learning, actually live — and every word of it is built on the single-variable shapes you just learned to read.

Problems

Six tiers. Fluency and Practice are doing* — classify critical points and solve real optimization problems until it’s automatic. Concept & counterexample and ◆ Diamond are thinking — the convexity ideas and the proofs. Teach it is teaching. Looking ahead opens Chapter 9. Do all of Fluency; the Concept tier here is where the convexity ideas actually take hold.*

Fluency — do these fast

1. Find all critical points of: (a) $f(x) = x^2 - 6x + 1$; (b) $f(x) = x^3 - 3x$; (c) $f(x) = x^4 - 2x^2$.
2. Use the second-derivative test to classify each critical point in problem 1.
3. State where each is concave up and concave down, and find any inflection points: (a) $f(x) = x^3$; (b) $f(x) = x^3 - 3x^2$.
4. Find the critical points of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$ and classify them. (Recall $f' = \cos x - \sin x$.)
5. For each, decide whether it’s convex, concave, or neither on all of $\mathbb{R}$, using the sign of $f''$: (a) $x^2$; (b) $e^x$; (c) $\ln x$ (on $x>0$); (d) $-x^2$; (e) $x^3$.
6. Find the global max and min of $f(x) = x^2 - 4x + 3$ on $[0, 3]$ using the closed-interval method.

Practice — apply and optimize

1. Find the global max and min of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.
2. A farmer has $200$ m of fencing and wants to enclose the largest possible rectangular area against a straight river (no fence needed on the river side). Find the dimensions. (Set up area as a function of one side, then optimize.)
3. Find the point on the line $y = 2x + 3$ closest to the origin. (Minimize the squared distance — it has the same minimizer and easier algebra.)
4. A box with a square base and open top must hold $32\,\text{cm}^3$. Find the dimensions using the least material. (Mirror the can example: express surface area in one variable, optimize, confirm with $f''$.)
5. Find the maximum of $f(x) = x\,e^{-x}$ for $x \ge 0$, and confirm it’s a maximum two ways (first- and second-derivative tests).
6. On $[0, \pi]$, find where $f(x) = \sin x\,(1 + \cos x)$ is largest. (A trig optimization — find $f'$, set it to zero.)

Concept & counterexample — think; true/false with justification, or build the example

1. 1. Every critical point is a local maximum or minimum. (b) If $f''(a) = 0$, then $a$ is an inflection point. (c) A convex function can have two distinct local minima. (d) If $f'(a) = 0$ and $f''(a) = 0$, the point is definitely not an extremum.
2. Find the error. A student says: “$f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$, so by the second-derivative test $x = 0$ is neither a max nor a min.” What’s wrong, and what is true at $x = 0$?
3. Construct a function with a local minimum that is not its global minimum. (This is exactly the “false floor” convexity forbids.)
4. Construct a critical point where $f'(a) = 0$ but $a$ is neither a max nor a min, and a different one where $f''(a) = 0$ yet $a$ is a minimum. (These show why each “inconclusive” case really is inconclusive.)
5. Is it possible? A convex function with a local maximum somewhere in the interior of its domain. Argue from the four faces why or why not.
6. Explain in two or three sentences why a convex loss function makes gradient descent guaranteed to find the best answer, while a non-convex one does not — referring to the two-valley picture.

◆ Diamond — the proofs are the foundation

1. ◆ Prove Fermat’s theorem (cite Fermat’s lemma from Chapter 6) and explain why the conclusion is only “$f'(c) = 0$ or $f'(c)$ doesn’t exist,” not just the first.
2. ◆ Prove the first-derivative test for a local maximum: if $f' > 0$ just left of $c$ and $f' < 0$ just right, then $c$ is a local max. (Use Chapter 6’s monotonicity consequence on each side.)
3. ◆ Prove the second-derivative test: if $f'(a) = 0$ and $f''(a) > 0$, then $a$ is a local minimum. (Use the Chapter 7 quadratic $f(x) \approx f(a) + \tfrac12 f''(a)(x-a)^2$, and argue the quadratic term dominates for $x$ near $a$.)
4. ◆ Prove that $f'' \ge 0$ implies the graph lies above each of its tangent lines (Face 4 $\Rightarrow$ Face 2), using the Taylor remainder. Then specialize to re-prove $e^x \ge 1 + x$.
5. ◆◆ Prove the keystone theorem: if $f$ is convex and $f'(a) = 0$, then $a$ is the global minimum. (Use the tangent-below face with the horizontal tangent at $a$.)
6. ◆◆ Prove Face 2 $\Rightarrow$ Face 1 (tangent below $\Rightarrow$ chord above): take two points, and show the chord between them stays on or above the graph. (Hint: a clean route is to first prove $f'' \ge 0 \Rightarrow$ for any $x$ between $p$ and $q$, $f(x)$ lies below the chord, using that $f'$ is increasing.)

Teach it — explain it, write it up, or fix it

1. In three or four sentences with the bowl-and-floor picture, teach a friend what convexity is and why it means “no false floors.”
2. Debug your classmate. They claim: “I found a point where $f'(x) = 0$, so it’s the minimum of $f$.” Explain in two sentences what they’ve actually found and the two things they still have to check before calling it a global minimum.
3. Write a complete worked solution to “find the rectangle of maximum area with perimeter $40$,” narrating each step — setup, the single-variable function, optimize, confirm — so a peer could redo it for any perimeter.
4. Design a problem. Invent an applied optimization problem (a box, a fence, a can, or a dose) whose answer is a clean number, and write the full solution key including the $f''$ check.
5. In a few sentences, teach the difference between a local and a global extremum, and explain why convexity is exactly the condition that makes the two coincide.

Looking ahead — Chapter 9 and Book 3 in your hands early

1. Gradient descent (Chapter 9) finds minima by stepping downhill until the slope is zero — i.e., it chases the equation $f'(x) = 0$ you’ve been solving by hand. For the convex function $f(x) = x^2$, predict: starting anywhere, will gradient descent find the global minimum? Why does convexity guarantee it?
2. For the non-convex $f(x) = x^4 - 3x^2 + x$ (the two-valley function), predict what determines which minimum gradient descent lands in, and connect this to why “where you start” mattered in the Chapter 9 lab.
3. In Book 3, $x$ becomes a vector and “$f' = 0$” becomes “the gradient is the zero vector.” Predict: what should replace the second derivative when there are many input directions, and what might “convex” mean for a bowl in many dimensions? (You’re guessing the shape of multivariable optimization — the real home of machine learning.)

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