Chapter 2 Review

Answer each question to the best of your ability before reading the solution.

What is the second derivative, how is it written, and what does its sign tell us?

Solution

The second derivative is the derivative of the first derivative:

$f''(x)=\dv{}{x}\bigl(f'(x)\bigr)=\dv[2]{y}{x}\text{.}$

It measures how $$ f' $$ changes. If $$ f''>0 $$ , then $$ f' $$ is increasing; if $$ f''<0 $$ , then $$ f' $$ is decreasing.
How are an object's position, velocity, and acceleration related? When is the object speeding up?

Solution

If $$ s(t) $$ is position, then

$v(t)=s'(t)\qand a(t)=v'(t)=s''(t)\text{.}$

The object speeds up when velocity and acceleration have the same sign, since acceleration reinforces the motion. It slows down when they have opposite signs.
What do marginal functions represent, and how can derivatives describe diminishing returns?

Solution
Marginal revenue, cost, and profit are the derivatives $$ R'(x) $$ , $$ C'(x) $$ , and $$ P'(x) $$ . Each estimates the change produced by one additional unit near $$ x $$ . Diminishing returns typically occur when $$ f'(x)>0 $$ but $$ f''(x)<0 $$ : the output is still increasing, but its rate of increase is decreasing.
What is a critical point, and why is it only a candidate for a local extremum?

Solution
A critical point is an input $$ c $$ in the domain of $$ f $$ where $$ f'(c)=0 $$ or $$ f'(c) $$ is undefined. The derivative might not change sign there, so the function may continue increasing or decreasing without having a local maximum or minimum.
State the First Derivative Test.
Solution
At a critical point $$ c $$ :
- A change in $$ f' $$ from positive to negative gives a local maximum.
- A change from negative to positive gives a local minimum.
- If $$ f' $$ has the same sign on both sides, there is no local extremum.
How does $$ f'' $$ determine concavity, and what establishes an inflection point?

Solution
If $$ f''>0 $$ , the graph is concave up; if $$ f''<0 $$ , it is concave down. An inflection point occurs where the concavity changes. Thus, $$ f''(c)=0 $$ or an undefined $$ f''(c) $$ provides only a candidate: the sign of $$ f'' $$ must change across $$ c $$ .
State the Second Derivative Test. When is it inconclusive?

Solution
If $$ f'(c)=0 $$ , then $$ f''(c)>0 $$ gives a local minimum and $$ f''(c)<0 $$ gives a local maximum. If $$ f''(c)=0 $$ or is undefined, the test is inconclusive; the First Derivative Test can still classify the point.
How can the graph of $$ f' $$ tell us about the shape of $$ f $$ ?

Solution
Where $$ f' $$ is above the horizontal axis, $$ f $$ increases; where it is below, $$ f $$ decreases. Crossings of the axis indicate local extrema when the sign changes. Where the graph of $$ f' $$ rises, $$ f $$ is concave up; where it falls, $$ f $$ is concave down.
What information is used to construct a qualitative sketch of a function?

Solution
Combine the domain, intercepts, boundary or long-run behavior, critical points, increasing and decreasing intervals, concavity, and inflection points. A qualitative sketch need not be drawn to scale, but its important points and its rising, falling, and bending behavior must agree with this information.
What are the objective, constraints, and feasible domain in an optimization problem?

Solution
The objective is the quantity to maximize or minimize. Constraints are relationships that feasible choices must satisfy. They allow us to rewrite the objective as a function of one variable. The feasible domain contains the inputs that satisfy the constraints and make sense in the original situation.
Distinguish a maximizer from a maximum value, and a local maximum from an absolute maximum.

Solution
A maximizer is an input $$ x^* $$ ; the maximum value is its output $$ f(x^*) $$ . A local maximum is at least as large as nearby outputs, while an absolute maximum is at least as large as every feasible output. The domain determines which comparisons are required and whether a best value is attained.
State the Absolute Extrema Test.

Solution
Every attained absolute maximum or minimum on an interval must occur at a feasible interior critical point or an included boundary point. These inputs are candidates, so we evaluate the objective at each one and compare the resulting outputs.
Explain how to solve an optimization problem. What common pitfalls should you watch for?

Solution
Identify the objective and constraints, reduce the objective to one variable, determine the feasible domain, find all critical and included boundary points, and compare the objective values. Reject infeasible results, report the requested quantity with units, and interpret it in context. Finally, remember that the result is optimal only within the assumptions of the model.
What is a tangent-line approximation, and why does it work near its base point?

Solution

The tangent line to $$ f $$ at $$ x=a $$ is

$L(x)=f(a)+f'(a)(x-a)\text{.}$

Near $$ a $$ , the function and line have nearly the same change, so $\Delta y\approx f'(a)\Delta x$ and $f(x)\approx L(x)$ . The approximation generally becomes less reliable farther from the base point.
How do concavity, error, and sensitivity help us judge or use a tangent-line approximation?

Solution

With consistent concavity, a concave-up graph lies above its tangent line, giving an underestimate; a concave-down graph gives an overestimate. Absolute error measures the difference in the original units, while relative error compares that difference with the actual value.

For tolerances, $\displaystyle |\Delta y|\approx|f'(a)|\,|\Delta x|$ . A large $$ |f'(a)| $$ means greater sensitivity and therefore requires tighter control of the input.