Calc I — Midterm 1 Zero-to-Exam Review

Worked: is it a function?

(a) \(y=x^2\) — every x gives one y ✓ (but note both \(x=2\) and \(x=-2\) give 4 — that's allowed, it's one output per input, not one input per output)

(b) \(x=y^2\) — try \(x=4\): \(y=2\) or \(y=-2\). One input, two outputs ✗ NOT a function.

(c) The circle \(x^2+y^2=9\) — vertical line through the middle hits twice ✗.

Worked: three domain examples

Example 1: \(f(x)=\dfrac{x+1}{x-2}\). Bottom zero at \(x=2\) → domain = all reals except 2. Written: \((-\infty,2)\cup(2,\infty)\) or "all \(x\neq 2\)".

Example 2: \(g(x)=\sqrt{x-5}\). Need \(x-5\ge0\) → \(x\ge 5\) → domain \([5,\infty)\).

Example 3: \(h(x)=\dfrac{\sqrt{x+3}}{x-1}\). Two conditions: \(x+3\ge0\) → \(x\ge-3\), AND \(x\neq1\). Domain: \([-3,1)\cup(1,\infty)\). Note the \(-3\) is INCLUDED (square root allows zero) but 1 is excluded (division can't be zero).

Worked: transforms

Example 1: The point \((2,6)\) is on \(f(x)=|x|\). Where does it land on \(f(x)=|4x|\)?
Inside change → horizontal, factor \(1/c=1/4\). x-coordinate gets divided by 4: new point \((\frac{2}{4},6)=(0.5,\,6)\). (This exact question style was on a past MC.)

Example 2: Describe \(g(x)=(x-3)^2+1\) from \(x^2\): shift right 3, up 1. Vertex moves from \((0,0)\) to \((3,1)\).

Example 3: \(h(x)=-2\sqrt{x}\): vertical stretch ×2 and flip upside down (the negative).

Worked: inverse, step by step

Find the inverse of \(f(x)=\dfrac{2x+3}{x-1}\).

1. Write \(y=\dfrac{2x+3}{x-1}\)
2. Swap: \(x=\dfrac{2y+3}{y-1}\)
3. Multiply both sides by \((y-1)\): \(x(y-1)=2y+3\)
4. Expand: \(xy-x=2y+3\)
5. Collect all y-terms on one side: \(xy-2y=x+3\)
6. Factor the y out: \(y(x-2)=x+3\)
7. Divide: \(y=\dfrac{x+3}{x-2}\)

So \(f^{-1}(x)=\dfrac{x+3}{x-2}\).

Verify (they love this as a T/F): \(f(f^{-1}(x))\) should equal \(x\): \(f\!\left(\frac{x+3}{x-2}\right)=\frac{2\frac{x+3}{x-2}+3}{\frac{x+3}{x-2}-1}=\frac{2(x+3)+3(x-2)}{x-2}=\frac{2x+6+3x-6}{x-2}=\frac{5x}{5}\)... hold on — let's recompute the bottom: \(\frac{x+3}{x-2}-1=\frac{x+3-(x-2)}{x-2}=\frac{5}{x-2}\). Top: \(\frac{2(x+3)}{x-2}+3=\frac{2x+6+3x-6}{x-2}=\frac{5x}{x-2}\). Ratio \(=\frac{5x/(x-2)}{5/(x-2)}=x\) ✓ Verified.

Worked: inverse of exponential/log

Example: \(f(x)=e^{2x}\). Inverse: swap → \(x=e^{2y}\) → \(\ln x = 2y\) → \(y=\frac{\ln x}{2}\). So \(f^{-1}(x)=\frac{\ln x}{2}\), domain \(x\gt0\).

Log laws that matter:

• \(\ln(ab)=\ln a+\ln b\)
• \(\ln\frac{a}{b}=\ln a-\ln b\)
• \(\ln(a^k)=k\ln a\)
• \(e^{\ln x}=x\) and \(\ln(e^x)=x\) — they undo each other

Worked: piecewise (both sides)

\(f(x)=\begin{cases}x+2 & x\lt 2\\ x^2 & x\ge 2\end{cases}\). Find \(\displaystyle\lim_{x\to 2}f(x)\).

1. LEFT side uses the top piece (x<2): \(\lim_{x\to 2^-}(x+2)=2+2=4\)
2. RIGHT side uses the bottom piece (x≥2): \(\lim_{x\to 2^+}(x^2)=4\)
3. 4 = 4 → the two-sided limit exists: \(\lim_{x\to2}f(x)=\boxed{4}\)
4. Note: \(f(2)=2^2=4\) here too, so \(f\) is actually continuous at 2 (more in section 4).

Worked: one side breaks → DNE

\(f(x)=\begin{cases}x+2 & x\lt 2\\ x^2+3 & x\ge 2\end{cases}\). Left: 4. Right: \(4+3=7\). \(4\neq7\) → jump → \(\lim_{x\to2}f(x)\) DNE. (Each one-sided limit exists fine — the two-sided one doesn't.)

The #1 exam pattern: 0/0 → factor → cancel → retry

When direct substitution gives \(\frac{0}{0}\), that number is FORBIDDEN but the limit still usually exists. Factor the top, kill the culprit, retry:

$$\lim_{x\to 3}\frac{x^2-9}{x-3}=\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to 3}(x+3)=\boxed{6}$$

Why cancelling is legal here: in the limit we only evaluate at \(x\neq3\), where \(\frac{x-3}{x-3}=1\) honestly. We never divide actual zero by actual zero.

Worked: the conjugate trick (square roots)

$$\lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}$$

Plug 0: \(\frac{3-3}{0}=\frac{0}{0}\). Can't factor a root, so multiply top AND bottom by the conjugate of the top \((\sqrt{x+9}+3)\):

1. \(\frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}\)
2. Top: \((\sqrt{x+9})^2-3^2=(x+9)-9=x\) (difference of squares!)
3. \(=\dfrac{x}{x(\sqrt{x+9}+3)}=\dfrac{1}{\sqrt{x+9}+3}\) (cancel the x)
4. Plug 0: \(\dfrac{1}{3+3}=\boxed{\dfrac16}\)

Worked: cube version (Stewart Example)

$$\lim_{x\to 2}\frac{x^3-8}{x-2}=\lim_{x\to 2}\frac{(x-2)(x^2+2x+4)}{x-2}=\lim_{x\to2}(x^2+2x+4)=4+4+4=\boxed{12}$$

Remember \(a^3-b^3=(a-b)(a^2+ab+b^2)\) from the warm-up? Told you factoring was the real skill.

Worked: direct substitution via the laws

$$\lim_{x\to2}\frac{x^3-2x+4}{x-1}$$

Bottom at 2: \(2-1=1\neq0\) → Quotient law is legal → evaluate top and bottom at 2: \(\frac{8-4+4}{1}=\boxed{8}\). One line, but you should NAME the legality: denominator ≠ 0 at 2.

Worked: a sneaky one — multiply out instead of factor

$$\lim_{x\to0}\frac{(3+x)^{-1}-(3)^{-1}}{x}=\lim_{x\to0}\frac{\frac{1}{x+3}-\frac13}{x}$$

0/0 again. Combine the top into ONE fraction: \(\frac{1}{x+3}-\frac13=\frac{3-(x+3)}{3(x+3)}=\frac{-x}{3(x+3)}\)

Then divide by x: \(\frac{-x}{3(x+3)}\cdot\frac1x = \frac{-1}{3(x+3)}\) → plug 0 → \(\boxed{-\frac19}\)

This is a real Stewart example. Pattern: complex fractions → combine to one fraction → cancel.

Worked: find k making it continuous (THE exam question)

$$f(x)=\begin{cases}\dfrac{x^2-4}{x-2} & x\neq2\\ k & x=2\end{cases} \quad\text{— find } k \text{ so } f \text{ is continuous everywhere.}$$

1. For \(x\neq2\): rational with no zero-bottom issues except at 2 → continuous there automatically.
2. Only question: at \(x=2\). Limit: \(\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}=\lim\frac{(x-2)(x+2)}{x-2}=4\)
3. Need \(f(2)=k\) to equal the limit → \(\boxed{k=4}\)

That's it. Limit exists (4), point exists (k), set them equal. Every "find k for continuity" problem is this recipe.

Worked: classify the break

\(g(x)=\begin{cases}x^2 & x\lt1\\ 3-x & x\ge1\end{cases}\). Continuous everywhere?

1. Both pieces are polynomials → continuous on their own territories.
2. At \(x=1\): left limit \(=1^2=1\); right limit \(=3-1=2\).
3. \(1\neq2\) → two-sided limit DNE → jump discontinuity at \(x=1\) — not removable.

Worked: the "continuous EXCEPT at" wording (directly from the corrected Q2b)

Say the practice file says: "\(f(x)\) is continuous on its domain except at \(x=-2\), where it is not continuous." What's being tested? You should be able to explain WHY it fails there — which of the 3 conditions dies. If \(f(-2)\) doesn't exist → condition 1. If sides disagree at −2 → condition 2. If \(f(-2)\) exists but disagrees with the limit → condition 3. Read the graph/table they give you and name the failed condition.

Worked proof 1 (linear — the standard one)

Prove: \(\lim_{x\to3}(2x+1)=7\).

1. Given ε>0. Want: \(|(2x+1)-7|\lt\varepsilon\).
2. Simplify: \(|2x-6|=2|x-3|\)
3. Need \(2|x-3|\lt\varepsilon\) ⟺ \(|x-3|\lt\varepsilon/2\)
4. Choose \(\boxed{\delta=\varepsilon/2}\). Then \(0\lt|x-3|\lt\delta\) implies \(|2x+1-7|=2|x-3|\lt2\delta=\varepsilon\). ∎

Worked proof 2 (quadratic — needs the "cap δ" trick)

Prove: \(\lim_{x\to2}x^2=4\).

1. Want: \(|x^2-4|=|x-2||x+2|\lt\varepsilon\)
2. Problem: the \(|x+2|\) factor floats around. Cap it: if \(|x-2|\le1\), then x∈[1,3] so \(|x+2|\le5\).
3. Then \(|x^2-4|\le5|x-2|\). Want ≤ ε ⟸ \(|x-2|\lt\varepsilon/5\).
4. Choose \(\delta=\min\{1,\ \varepsilon/5\}\) (min = "whichever is smaller, so BOTH conditions hold"). ∎

The ε–δ lecture video the prof posted (Azar) walks exactly this shape. If a proof question appears, it will be linear like Proof 1 — quadratic if they're feeling mean.

Worked: one of each

a) \(\displaystyle\lim_{x\to\infty}\frac{4x-1}{x^3+2}=\) top degree 1 < bottom degree 3 → \(\boxed{0}\)

b) \(\displaystyle\lim_{x\to\infty}\frac{3x^2-5x+1}{5x^2+x}=\) tie at degree 2 → \(\frac{3}{5}\)

c) \(\displaystyle\lim_{x\to\infty}\frac{x^2+1}{x-3}=\) top wins → grows like x → \(+\infty\) → DNE (no HA)

Worked: the divide-by-highest-power method (show the work!)

$$\lim_{x\to\infty}\frac{3x^2-5x+1}{5x^2+x}$$

1. Divide every term, top and bottom, by \(x^2\): \(\dfrac{3-5/x+1/x^2}{5+1/x}\)
2. As \(x\to\infty\): each \(\frac{1}{x}\) and \(\frac{1}{x^2}\) → 0 (limits at infinity of tiny-over-huge are 0)
3. \(=\dfrac{3-0+0}{5+0}=\boxed{\dfrac35}\)

Write it this way on the exam — the table above is for speed-checking, the division is for marks.

Worked: root flavor

$$\lim_{x\to\infty}\frac{\sqrt{x^2+9}}{2x+3}$$

1. \(\sqrt{x^2+9}=\sqrt{x^2(1+9/x^2)}=\sqrt{x^2}\cdot\sqrt{1+9/x^2}\). For \(x\to+\infty\), \(\sqrt{x^2}=x\) → \(=x\sqrt{1+9/x^2}\)
2. \(\dfrac{x\sqrt{1+9/x^2}}{2x+3}=\dfrac{\sqrt{1+9/x^2}}{2+3/x}\to\dfrac{\sqrt{1+0}}{2}=\boxed{\dfrac12}\)

Careful: for \(x\to-\infty\), \(\sqrt{x^2}=\textbf{|x|}=-x\) — sign flips! This asymmetry is a classic trap.

Worked: prove a root exists (the classic, verbatim format)

Show that \(x^3+x-4=0\) has a solution in \((1,2)\).

1. Let \(f(x)=x^3+x-4\). It's a polynomial → continuous everywhere, in particular on \([1,2]\). (Say this — it's a mark.)
2. Evaluate endpoints: \(f(1)=1+1-4=-2\) (negative); \(f(2)=8+2-4=6\) (positive)
3. \(N=0\) is between \(-2\) and 6
4. IVT ⇒ there exists \(c\in(1,2)\) with \(f(c)=0\), i.e., \(c^3+c-4=0\). ∎ A root exists in \((1,2)\).

They can also ask "how many times can it cross?" — that's IVT + a monotonicity argument, usually later material. For the midterm: existence is the ask.

Worked 1: quadratic from scratch (master this one cold)

\(f(x)=x^2+3x\). Find \(f'(x)\) using the definition (they will demand the definition, not the shortcut).

1. Need \(f(x+h)\): replace every x with (x+h): \(f(x+h)=(x+h)^2+3(x+h)=x^2+2xh+h^2+3x+3h\)
2. Difference: \(f(x+h)-f(x)=(x^2+2xh+h^2+3x+3h)-(x^2+3x)=2xh+h^2+3h\)
3. Divide by h: \(\frac{h(2x+h+3)}{h}=2x+h+3\) for \(h\neq0\)
4. Send \(h\to0\): \(\boxed{f'(x)=2x+3}\)

Step 1 is where everyone dies. Practice writing \((x+h)^2=x^2+2xh+h^2\) until automatic.

Worked 2: fraction flavor

\(f(x)=\dfrac1x\). Find \(f'(x)\) by definition.

1. \(f(x+h)-f(x)=\frac1{x+h}-\frac1x=\frac{x-(x+h)}{x(x+h)}=\frac{-h}{x(x+h)}\)
2. Divide by h: \(\frac{-h}{x(x+h)}\cdot\frac1h=-\frac{1}{x(x+h)}\)
3. \(h\to0\): \(\boxed{f'(x)=-\frac{1}{x^2}}\)

Worked 3: root flavor (conjugate again!)

\(f(x)=\sqrt{x}\). Find \(f'(x)\) by definition.

1. \(\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)
2. \(h\to0\): \(\boxed{f'(x)=\frac{1}{2\sqrt{x}}}\)

Spot the pattern — conjugate trick shows up in BOTH limits and derivatives. One trick, double duty.

Worked 4: tangent line equation (very common Q)

Find the tangent line to \(f(x)=x^2+3x\) at \(x=1\) (using our result \(f'(x)=2x+3\)).

1. Point: \(f(1)=1+3=4\) → point \((1,4)\)
2. Slope: \(f'(1)=2(1)+3=5\)
3. Point-slope: \(y-4=5(x-1)\) → \(y=5x-1\)

Format: point (plug into f), slope (plug into f'), then \(y-y_1=m(x-x_1)\). Memorize that 3-beat.

Worked: negative + fractional powers (where marks die)

a) \(f(x)=\dfrac{1}{x}\) — rewrite as \(x^{-1}\) → \(f'(x)=-x^{-2}=-\dfrac{1}{x^2}\) (matches the definition answer from section 8 — sanity check!)

b) \(f(x)=\sqrt{x}\) — rewrite as \(x^{1/2}\) → \(f'(x)=\frac12x^{-1/2}=\dfrac{1}{2\sqrt{x}}\) ✓

c) \(f(x)=\sqrt[3]{x^2}=x^{2/3}\) → \(f'(x)=\frac{2}{3}x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}\)

THE move: before anything else, rewrite every root and every 1/(...) into exponent form. Then differentiate. Then rewrite back pretty.

Worked: polynomials end-to-end

\(f(x)=x^5-4x^3+6x-\pi\)

\(f'(x)=5x^4-12x^2+6-0\) (π is a constant → derivative 0)

Worked: product rule ×2

a) \(f(x)=x^2\sin x\): \(f'g+fg'=(2x)(\sin x)+(x^2)(\cos x)=\boxed{2x\sin x+x^2\cos x}\)

b) \(f(x)=(x^2+1)(x^3-3x)\): two routes —

• Product rule: \((2x)(x^3-3x)+(x^2+1)(3x^2-3)=2x^4-6x^2+3x^4-3x^2+3x^2-3=5x^4-6x^2-3\)
• Expand first: \(x^5-x^3+3x^2-3\) → \(5x^4-3x^2+6x\)... careful — expand: \((x^2+1)(x^3-3x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x\) → derivative \(5x^4-6x^2-3\) ✓ same. Good — cross-check your product rule with expansion whenever you can.

Worked: quotient rule ×2

a) \(y=\dfrac{x}{x^2+1}\): \(\dfrac{(1)(x^2+1)-(x)(2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(1+x^2)^2}\)

b) \(y=\dfrac{x^2+2x}{x+1}\): \(\dfrac{(2x+2)(x+1)-(x^2+2x)(1)}{(x+1)^2}=\dfrac{2(x+1)^2-x^2-2x}{(x+1)^2}=\dfrac{2x^2+4x+2-x^2-2x}{(x+1)^2}=\dfrac{x^2+2x+2}{(x+1)^2}\)

Worked: proving the tan rule (this exact proof was a lecture video)

Derive \((\tan x)'=\sec^2x\) from sin/cos:

1. \(\tan x=\dfrac{\sin x}{\cos x}\) → quotient rule: \(\dfrac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^2x}\)
2. \(=\dfrac{\cos^2x+\sin^2x}{\cos^2x}\)
3. Numerator = 1 (identity!) → \(=\dfrac{1}{\cos^2x}=\sec^2x\) ∎

Worked: sec by quotient rule (prof's video example)

\(\sec x=(\cos x)^{-1}\)

1. Quotient/power: \(y'=-(\cos x)^{-2}\cdot(-\sin x)=\dfrac{\sin x}{\cos^2x}\)
2. \(=\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}=\sec x\tan x\) ∎

Worked: mixed differentiation

a) \(f(x)=3\sin x-4\cos x+2\tan x\) → \(f'(x)=3\cos x+4\sin x+2\sec^2x\)

b) \(f(x)=x\cos x\) (product!) → \(f'(x)=\cos x-x\sin x\)

c) \(f(x)=\dfrac{\sin x}{x}\) (quotient!) → \(f'(x)=\dfrac{x\cos x-\sin x\cdot1}{x^2}=\dfrac{x\cos x-\sin x}{x^2}\)

The limit-proof they filmed (sine from first principles)

\(\lim_{h\to0}\frac{\sin h}{h}=1\) and \(\lim_{h\to0}\frac{\cos h-1}{h}=0\) — these two facts + squeeze theorem are how Stewart proves \((\sin x)'=\cos x\). You won't redo the full proof, but if asked where \((\sin x)'=\cos x\) comes from: "two special limits + the definition."

Worked: chain ×3

a) \(\sin(3x)\): outside sin, inside 3x → \(\cos(3x)\cdot3=\boxed{3\cos(3x)}\)

b) \((x^2+1)^5\): outside (·)⁵, inside \(x^2+1\) → \(5(x^2+1)^4\cdot2x=\boxed{10x(x^2+1)^4}\)

c) \(\sqrt{x^2+4}\): outside √ → \( (x^2+4)^{1/2}\) → \(\frac12(x^2+4)^{-1/2}\cdot2x=\boxed{\dfrac{x}{\sqrt{x^2+4}}}\)

Worked: circle slope

\(x^2+y^2=25\). Find \(\frac{dy}{dx}\).

1. \(\frac{d}{dx}(x^2)=2x\); \(\frac{d}{dx}(y^2)=2y\cdot\frac{dy}{dx}\) (y is an inside function — chain!); \(\frac{d}{dx}(25)=0\)
2. \(2x+2y\dfrac{dy}{dx}=0\)
3. \(\dfrac{dy}{dx}=-\dfrac{x}{y}\)

Sanity: at the top of the circle (0,5) slope = 0 ✓ (flat); at the side (5,0) it's vertical ✓. The formula knows the circle.

Worked: tangent via implicit

Find the tangent to \(x^2+xy+y^2=7\) at \((1,2)\).

1. Differentiate: \(2x+y+xy'+2yy'=0\) (product rule on the xy term: \(1\cdot y+x\cdot y'\))
2. Plug point: \(2+2+y'+4y'=0\) → \(4+5y'=0\) → \(y'=-4/5\)
3. Tangent: \(y-2=-\frac45(x-1)\)