MATH 116 Test 2 Practice (Fall 2024)

You need the derivative of a product \(R(x) = x\cdot p(x)\).

Look up the product rule for derivatives.

1. First write \(R(x)\) explicitly: \[ R(x) = x(800-0.2x). \]
2. Either expand and use the power rule, or apply the product rule directly.
3. After finding \(R'(x)\), plug in \(x=1600\).

Keep track of units: revenue is in dollars per unit of \(x\).

1. First find \(q'(p) = \dfrac{dq}{dp}\) using the power rule.
2. Then plug into the elasticity formula \[ E(p) = -\frac{p}{q(p)}\,q'(p). \]

1. Compute \(q(30)\).
2. Compute \(q'(30)\).
3. Then calculate \[ E(30) = -\frac{30}{q(30)}\cdot q'(30) \] and approximate as a decimal.
4. Compare your value to the intervals in the choices.

This is a product of two functions: \(x^{2}\) and \(f(x)\).
Use the product rule.

The product rule says:

1. Compute \((x^{2})'\).
2. Plug in \(x=3\), \(f(3)=-2\), and \(f'(3)=5\).

\(h(x)\) is a quotient of two differentiable functions.
Use the quotient rule.

Quotient rule:

1. Plug in \(x=2\).
2. Carefully substitute the four given values: \(f(2), f'(2), g(2), g'(2)\).
3. Simplify the fraction.

Marginal cost \(C'(x)\) is the instantaneous rate of change of cost with respect to units produced.

It is measured in dollars per additional bottle.

Think: When \(x=8\) bottles, what does “\(C'(8) = 20\)” say about how total cost changes if we increase production slightly?

Your response should mention “approximately” and talk about the next bottle (or next few bottles).

Rewrite the square root as a power and use the chain rule.

Write

• Outside function: \(f(u)=u^{1/2}\).
• Inside function: \(u(x) = -5x^{2}+7x\).

Use

Compute both \(f'(u)\) and \(u'(x)\).

Elasticity key idea:

• If \(|E(p)| > 1\): elastic.
• If \(|E(p)| < 1\): inelastic.

Inelastic demand behaves differently for revenue than elastic demand.

Here \(E(p) = 0.84\).

1. Decide if that is elastic or inelastic.
2. Recall:
   • With elastic demand, increasing price makes revenue go down.
   • With inelastic demand, increasing price makes revenue go up.

Relative extrema occur at critical numbers (where \(f'(x)=0\) or undefined).

Solve \(f'(x)=0\) and then use a sign chart or second derivative to classify each critical number.

1. Factor or use the quadratic formula to solve \[ 2x^{2} + 6x - 20 = 0. \]
2. For each critical number, pick test values on the left and right to see whether \(f'\) changes from + to − (max) or − to + (min).

To find where a function is increasing, look at where its derivative is positive.

So:

1. Find \(P(x)\).
2. Compute \(P'(x)\).
3. Solve \(P'(x) > 0\).

1. Subtract \(C(x)\) from \(R(x)\) to form \(P(x)\) and simplify.
2. Differentiate the resulting polynomial.
3. Factor \(P'(x)\) and find critical points.
4. Use a sign chart to find which intervals make \(P'(x)>0\).

On the graph, where is \(f'(x)\) below the x-axis?

Those x-values are where \(f'(x) < 0\).

1. Identify the x-values where the graph of \(f'(x)\) crosses the x-axis.
2. Check visually which regions are below the x-axis.
3. Express that region as an open interval (or union of intervals).

(From the original test, the negative region is approximately from \(x=2\) to \(x=6\).)

Compute the derivative \(g'(x)\) using the power rule, then solve \(g'(x) = 0\).

1. Differentiate: \[ g'(x) = (x^{3})' - 6(x^{2})'. \]
2. Factor out common terms in \(g'(x)\).
3. Use the zero product property to find the critical numbers.

• Use the power rule to find \(f'(x)\).
• The tangent line at \(x=a\) uses the point-slope formula \(y - f(a) = f'(a)(x-a)\).
• For the First Derivative Test, find where \(f'(x)\) changes sign.

1. Differentiate \(2x^{3} - 9x^{2} - 86\).
2. Evaluate both \(f(1)\) and \(f'(1)\).
3. For increasing/decreasing:
   • Solve \(f'(x) = 0\) to get critical numbers.
   • Create a sign chart for \(f'(x)\).
   • Positive derivative ⇒ increasing; negative derivative ⇒ decreasing.

• For (a), rewrite powers: \(x^{-2}\) and \(x^{1/2}\). Use the power rule.
• For (b), use the chain rule on \([\text{inside}]^{7}\).
• For (c), use the product rule.

(a) Rewrite:

(b) For

use

(c) Let

then

Marginal cost is the derivative \(C'(x)\).

Since \(C(x)\) is a ratio of two functions, use the quotient rule.

Let

Then

After finding \(C'(x)\), evaluate at \(x=20\) and remember that the cost is in thousands of dollars, so the derivative will be in thousands of dollars per unit.

We are given

and the revenue function

First expand:

Now differentiate term–by–term:

Now evaluate at \(x=1600\):

So the marginal revenue at 1600 units is $160 per unit, which is more than $155.

Correct choice: (A).

We have

Step 1: Compute \(q'(p)\)

Step 2: Evaluate at \(p=30\)

Step 3: Elasticity

Elasticity of demand:

At \(p=30\):

So \(E(30)\) is between \(1.0\) and \(1.2\).

Correct choice: (C).

We are given

with \(f(3) = -2\) and \(f'(3) = 5\).

Use the product rule with

Then

Evaluate at \(x=3\):

Correct choice: (D).

We have

and the values

Using the quotient rule:

Evaluate at \(x=2\):

Correct choice: (B).

\(C(x)\) is total cost in dollars to produce \(x\) water bottles.
\(C'(x)\) is marginal cost: the instantaneous rate of change of cost with respect to quantity.

Given

this means:

When 8 bottles are produced, the cost is increasing at approximately $20 per additional bottle.

So producing the 9th bottle costs about $20 more than producing 8 bottles.

This matches statement (A).

We have

Let

Then \(g(x) = u(x)^{1/2}\).

By the chain rule,

Compute \(u'(x)\):

So

Correct choice: (A).

Elasticity at this price is

Since \(|E(p)| = 0.84 < 1\), demand is inelastic.

For inelastic demand:

• When price increases, the percentage drop in quantity is smaller than the percentage increase in price, so revenue increases.

Therefore the correct interpretation is:

The demand is inelastic, so as price increases, revenue increases.

Correct choice: (D).

We have

Set \(f'(x)=0\) to find critical numbers:

Factor:

so

Make a sign chart for

• For \(x<-5\), both factors negative ⇒ product positive ⇒ \(f'(x)>0\) (increasing).
• For \(-5
• For \(x>2\), both positive ⇒ \(f'(x)>0\) (increasing).

So:

• At \(x=-5\), derivative changes \(+\to-\) ⇒ relative maximum.
• At \(x=2\), derivative changes \(-\to+\) ⇒ relative minimum.

Correct choice: (A).

Revenue and cost:

Profit:

Differentiate:

Critical numbers:

On \(00\) ⇒ \(P'(x)>0\).
On \(x>6100\), that factor is negative ⇒ \(P'(x)<0\).

So profit is increasing exactly on \((0,6100)\).

Correct choice: (B).

We are given the graph of \(y=f'(x)\).

Where is \(f'(x) < 0\)? Wherever the graph lies below the x-axis.

From the picture, \(f'(x)\) crosses the x-axis around \(x=2\) and \(x=6\) and is below the axis between those two points.

So:

Correct choice: (B).

Differentiate:

Solve \(g'(x)=0\):

Derivative exists everywhere, so the critical numbers are \(x=0\) and \(x=4\).

Correct choice: (C).

We have

(a) Derivative

(b) Tangent line at \(x=1\)

Point: \((1,-93)\); slope: \(-12\).

Tangent line:

(c) First Derivative Test

Critical numbers from \(f'(x)=6x(x-3)=0\):

Sign chart:

• For \(x<0\): \(f'(x)>0\) ⇒ increasing.
• For \(0
• For \(x>3\): \(f'(x)>0\) ⇒ increasing.

So:

• Increasing on \((-\infty,0)\cup(3,\infty)\).
• Decreasing on \((0,3)\).

(a) \(f(x) = \dfrac{3}{x^{2}} - 4\sqrt{x}\)

Rewrite:

Differentiate:

(b) \(g(x) = (-x^{3} + 7x^{2} - 3)^{7}\)

Let \(u(x)=-x^{3}+7x^{2}-3\). Then \(g(x)=u(x)^{7}\).

By the chain rule:

(c) \(h(x) = (5x^{2} + 3)(4x+2)\)

Let \(f_1(x)=5x^{2}+3\), \(f_2(x)=4x+2\).

Product rule:

in thousands of dollars.

Let \(f(x)=8x+3\), \(g(x)=5x+7\). Then \(C(x) = f(x)/g(x)\).

Quotient rule:

Simplify numerator:

So

At \(x=20\):

Units: thousands of dollars per unit.

So marginal cost in dollars is

Rounded: