Math 135: Toolkit 🧰

1. $\dfrac{a}{b}\cdot \dfrac{c}{d} = \dfrac{ac}{bd}$
   
   English Multiplying fractions requires multiplying global context of the numerators and global context of the denominators.
2. $\dfrac{a}{b}\div \dfrac{c}{d} = \dfrac{a}{b}\cdot \dfrac{d}{c}$
   
   English Dividing fractions requires taking the reciprocal of the right fraction, then multiplying.
3. $\dfrac{a}{c} + \dfrac{b}{c} = \dfrac{a + b}{c}$
   
   English Adding fractions with the same denominator requires adding the global context of the numerators together.
4. $\dfrac{a}{b} + \dfrac{c}{d}\qquad $ Find the LCD (getting the same denominator), then use Property 3.
5. $\dfrac{ac}{bc} = \dfrac{a}{b}$
   English Cancelling an expression $c$ requires the expression to be a global factor in both the numerator and denominator.

Let $a, b, c$ be real numbers.

• Commutative Properties
  • $a + b = b + a$
  • $ab = ba$
  English Can swap terms with terms and factors with factors.
• Associative Properties
  • $(a + b) + c = a + (b + c)$
  • $(ab)c = a(bc)$
  English Can swap evaluation priority among all terms or among all factors.
• Distributive Property
  • $a(b + c) = ab + ac$
  • $(b+c)a = ab + ac$
  English The only way factors and terms interact.

1. $-a = (-1)a$
2. $-(-a) = a$
Tip Even number of negatives is positive.

1. $(A + B)(A - B) = A^2 - B^2$
2. $(A + B)^2 = A^2 + 2AB + B^2$
3. $(A - B)^2 = A^2 - 2AB + B^2$

A curve in the plane is the graph of a function if no vertical line intersects the curve more than once.

Imagine squishing the graph towards the $x$-axis:

When finding the difference quotient, the global structure will (after some algebra) be global factors \[\dfrac{h\cdot (\text{stuff})}{h}\] where the $h$ cancels.

• Domain: $\mathbb{R}$
• Range: depends (on the degree).
• Graph is smooth (no corners or sharp corners) and continuous (no breaks or holes).

• Common logarithm (base 10): $\log(x) = \log_{10}(x)$
• Natural logarithm (base $e$): $\ln (x) = \log_e(x)$

Let $f(x)$ be differentiable on the interval $(a, b)$.

• If $f'(x) > 0$ for all $x$ in $(a, b)$, then $f(x)$ is increasing on $(a, b)$.
• If $f'(x) < 0$ for all $x$ in $(a, b)$, then $f(x)$ is decreasing on $(a, b)$.

Given a function $f(x)$, follow these steps to determine the intervals on which $f(x)$ is increasing or decreasing:

1. Find the derivative of $f(x)$.
2. Find the critical numbers of $f(x)$, meaning:
   • Find the $x$-coordinates where $f'(x) = 0$ or $f'(x)$ is undefined (DNE)
3. Draw a sign diagram of $f'(x)$, and determine the sign of $f'(x)$ at one test value in each of the intervals created by critical numbers.
4. Use the Increasing/Decreasing Test to decide what intervals $f(x)$ is increasing/decreasing on.

Let $f(x)$ be a function where $f''(x)$ exists on $(a, b)$.

• If $f''(x) > 0$ for all $x$ in $(a, b)$, then the graph of $f(x)$ is concave upward on $(a, b)$.
• If $f''(x) < 0$ for all $x$ in $(a, b)$, then the graph of $f(x)$ is concave downward on $(a, b)$.

Given a function $f(x)$, follow these steps to determine the intervals of concavity:

1. Find the second derivative of $f(x)$.
2. Find the inflection points of $f(x)$, meaning:
   • Find the $x$-coordinates where $f''(x) = 0$ or $f''(x)$ is undefined (DNE)
3. Draw a sign diagram of $f''(x)$, and determine the sign of $f''(x)$ at one test value in each of the intervals created by $x$-coordinates of inflection points.
4. Use the Concavity Test to determine the intervals of concavity for $f(x)$.

Given a function $f(x)$, use this test to find the local extrema of $f(x)$:

1. Find the derivative of $f(x)$.
2. Find the critical numbers of $f(x)$.
3. For each critical number, examine the sign of $f'(x)$ to the left and right of $c$:
   1. If $f'(x)$ changes from positive to negative at $x = c$, then $f(x)$ has a local maximum at $(c, f(c))$.
   2. If $f'(x)$ changes from negative to positive at $x = c$, then $f(x)$ has a local minimum at $(c, f(c))$.
   3. If $f'(x)$ does not change sign at $x = c$, then $(c, f(c))$ is neither a local minimum nor maximum.

Suppose that $f'(c) = 0$, and $f''(x)$ exists on an interval containing $c$.

1. If $f''(c) > 0$, then $f(c)$ is a local minimum.
2. If $f''(c) < 0$, then $f(c)$ is a local maximum.
3. If $f''(c) = 0$, then the test fails. In this case, use the First Derivative Test on the critical number $c$.

Suppose $f$ is a continuous function defined on a closed interval $[a, b]$. Then $f$ is guaranteed to attain a global maximum and global minimum.

Given a continuous function $f$ defined on a closed interval $[a, b]$, use this method to find global extrema.

1. Find the critical numbers of $f$ in the open interval $(a, b)$.
2. Evaluate $f$ at each of the critical numbers.
3. Evaluate $f$ at each endpoint (meaning, find $f(a)$ and $f(b)$).
4. The least of these is the global minimum, and the greatest is the global maximum.

Suppose that $c$ is the only critical number of $f(x)$.

1. If $f''(x) > 0$ on the domain of $f(x)$, then $f(c)$ is both a local and global minimum.
2. If $f''(x) < 0$ on the domain of $f(x)$, then $f(c)$ is both a local and global maximum.

Since $P = R - C$, differentiating gives $P' = R' - C'$. Setting $P' = 0$ gives \[R'(q) = C'(q)\] Profit is maximized when marginal revenue equals marginal cost.

If the price increases by 1%, the demand will decrease by $E$%.

Note: The demand function must be written as $q = f(p)$ (quantity as a function of price) to calculate elasticity.

$E$ is a dimensionless quantity: \[E = -\dfrac{p}{q} \cdot \underbrace{\dfrac{dq}{dp}}_{\text{units/\$}} = \dfrac{\%\text{ change in quantity demanded}}{\%\text{ change in price}}\]

The derivative $y'$ from implicit differentiation gives the slope of the tangent line at a point $(a, b)$ on the curve. Plug in both $x = a$ and $y = b$ to evaluate the slope.

Each symbol in $\displaystyle\int_a^b f(x)\, dx$ represents an idea from the Riemann sum:

• $\displaystyle\int$ means "add up smaller and smaller width rectangles"
• $f(x)$ is the height of one rectangle
• $dx$ is the width of one rectangle
• $a$ and $b$ are the bounds of the interval $[a, b]$

The definite integral $\displaystyle\int_a^b f(x)\, dx$ computes signed area between the curve of $f(x)$ and the horizontal axis:

• If $f(x)$ is above the horizontal axis, the area is positive.
• If $f(x)$ is below the horizontal axis, the area is negative.

If $f(x)$ represents a rate of change, then the area underneath the curve (the definite integral) represents the total amount of something.

All particular antiderivatives of $f(x)$ differ only by a constant. The general antiderivative of $f(x)$ is \[F(x) + C\] where $C$ is an arbitrary constant.

If $f(t)$ is continuous on $[a, b]$, then the function \[g(x) = \int^x_a f(t) \ dt, \qquad a \leq x \leq b\] is also continuous on $[a, b]$, differentiable on $(a, b)$, and $g'(x) = f(x)$.

• The integral and derivative are inverse processes; each undoes what the other has done.
• The rate of change of the area underneath the curve $f(t)$ with respect to the right bound of i ntegration is the value of the function at the right boundary.
• This theorem guarantees the existence of an antiderivative of $f(t)$, which is $g(x)$.

If $f(x)$ is continuous on $[a, b]$, then \[\int^b_a f(x) \ dx = F(b) - F(a)\] where $F(x)$ is any antiderivative of $f(x)$.

If $F'(x)$ is a rate of change, then \[\int^b_a F'(x) = F(b) - F(a)\] gives the net change of $F(x)$ on $[a, b]$.

When integrating a function that involves a composition $f(g(x))$, set $u = g(x)$, or the inside of the composition. The differential of $u$ is $du = g'(x) \ dx$. We substitute: \[\int f'(g(x))\cdot g'(x)\ dx = \int f'(u)\ du\]

When using $u$-substitution on a definite integral, convert the bounds of integration from $x$-values to $u$-values. No back-substitution is needed: \[\int^b_a f'(g(x)) \cdot g'(x) \, dx = \int^{g(b)}_{g(a)} f'(u) \, du\]

1. Generally, $u$ is the inside of the composition. If you can identify any composition, try picking $u$ to be the inside function first.
2. With a fraction, try the denominator as $u$ and check if $du$ matches the numerator.
3. Try multiple $u$'s. You will know you picked the right $u$ when you can replace $x$ and $dx$ entirely with $u$'s.

Given an integrand with two factors, choose $u$ and $dv$. Then: \[\int u \, dv = uv - \int v \, du\] For definite integrals: \[\int^b_a u \, dv = uv \bigg\rvert^b_a - \int^b_a v \, du\]

From your choice of $u$, you find $du$ (differentiate). From your choice of $dv$, you find $v$ (integrate).

1. Try to choose $u$ that simplifies when differentiated.
2. Try to choose $dv$ where either:
   • $dv$ has a basic integration rule when integrated, or
   • $dv$ simplifies when integrated.

1. $\displaystyle\int \dfrac{1}{x^2 - a^2} \, dx = \dfrac{1}{2a} \ln\left\lvert \dfrac{x - a}{x + a} \right\rvert + C$
2. $\displaystyle\int \dfrac{1}{\sqrt{x^2 + a^2}} \, dx = \ln\left\lvert x + \sqrt{x^2 + a^2} \right\rvert + C$

Let $n$ be an integer. Use this table to identify which method is appropriate.

General advice:

• If $x^n$ and $\ln(x)$ are present as factors, use IBP with $u = \ln(x)$.
• If $x^n$ and $e^x$ are present as factors, use IBP with $u = x^n$.
• If the integrand is a composition (inside/outside structure), try $u$-sub first.

Always set up the integral as $\displaystyle\int (\text{top} - \text{bottom}) \, dx$. Look at the graph to determine which function is on top.

• This area is never negative, even if both $f(x)$, $g(x)$ are below the $x$-axis.
• If the curves cross on $[a, b]$, split the integral at the crossing point(s), and set up each piece with the correct top and bottom.
• This computes net geometric area (always $\geq 0$). In contrast, $\int^b_a f(x)\ dx$ computes net signed area between the curve and the $x$-axis.

Consumer surplus measures "how much of a deal you got" at the aggregate level. Both CS and PS are areas between two curves on $[0, q^*]$, computed as $\displaystyle\int (\text{top} - \text{bottom}) \, dq$.

A dollar today is worth more than a dollar in the future, because today's dollar could be invested and grow.

Present value answers: What is the lump sum payment required right now to match the value of the income stream over $T$ years?