MATH 116 Test 3 Practice (Fall 2024)

You have \(\ln(\text{something})\), not just \(\ln x\).
Look for the rule for the derivative of \(\ln(f(x))\).

Use

Here \(f(x)\) (inside the log) is \(5x+7\).
Compute its derivative and plug into the formula.

We want

Let the inside function be

Then

Compute \(u'(x)\):

So

Answer: A

The bases on both sides are exponentials with base \(e\).
Try rewriting the right-hand side as \(e^{(\text{something})}\) so you can equate exponents.

Use the fact that

So

Then set exponents equal and solve the resulting linear equation in \(x\).

Start with

Rewrite the right side using \((a^{b})^{c} = a^{bc}\):

Now we have

Because the exponential function \(e^{(\cdot)}\) is one-to-one,

Solve for \(x\):

Compare with the intervals:

• \(2\) is between \(1.5\) and \(3\).

So the solution lies in

Answer: C

You are given a rate of change \(f'(t)\) and an initial value \(f(0)\).
To recover \(f(t)\), integrate \(f'(t)\) and then use the initial value.

1. Compute \[ f(t) = \int (5.6t - 7.5)\,dt + C. \]
2. Plug in \(t=0\) and \(f(0)=20\) to find \(C\).
3. Evaluate \(f(6)\) for the year 2006 (since 2006 is 6 years after 2000).

We know:

• \(f'(t) = 5.6t - 7.5\),
• \(f(0) = 20\).

Step 1: Integrate \(f'(t)\)

Step 2: Use initial condition \(f(0)=20\)

So

Step 3: Evaluate at \(t=6\) (year 2006)

So there were approximately 75.8 billion monthly text messages in 2006.

This lies between 70 and 80 billion.

Answer: B

To get the equation of the tangent line to a point you need a slope and apoint the line goes trough. How can You use the derivative to find the slope you need? How can you find the point you need?

We have

Step 1: Find the point on the curve at \(x=0\)

So the point is \((0,5)\).

Step 2: Find the derivative

Differentiate term by term:

• For \(e^{-7x}\), use chain rule:
  \[ \frac{d}{dx}\bigl(e^{-7x}\bigr) = e^{-7x}\cdot(-7) = -7e^{-7x}. \]
• For \(4\), derivative is \(0\).

So

Step 3: Evaluate slope at \(x=0\)

Step 4: Use point-slope form

So the correct tangent line is

Use the relationship between effective rate \(E\) and nominal rate \(r\) for continuous compounding:

Plug \(E=0.106\) into

solve for \(r\), then convert to a percentage and see which interval fits.

We are given an effective rate of \(E = 10.6\% = 0.106\). For continuous compounding,

So

Approximate:

• \(\ln(1.1) \approx 0.0953\),
• \(\ln(1.106)\) is slightly larger, about \(0.1008\).

So

This lies between 9.90% and 10.20%.

Answer: B

You’re going backwards from derivative to function.
This is an indefinite integral / antiderivative problem:

Use the power rule for integrals:

Apply it to \(10x\) and to the constant \(2\) separately.

We want a function \(f(x)\) whose derivative is

Compute the antiderivative:

Compare with the options:

• Option B is exactly \(f(x) = 5x^{2} + 2x + C\).

So

Answer: B

Split the integral into two parts:

Use:

• exponential rule for \(e^{4x}\),
• power rule for \(x^{-3}\).

Recall:

and for the power rule with \(n\neq -1\),

Apply with \(n=-3\).

We have

First integral:

Second integral:

Combine them (one constant is fine):

You have a quotient: numerator \(x\), denominator \(3+\ln x\).
Use the quotient rule.

Let

Then

Compute \(N'(x)\) and \(D'(x)\) first.

Write

Let

Compute derivatives:

By the quotient rule,

First find an antiderivative \(F(x)\) of \(5x^{4}\), then use

Use the power rule:

Here \(5x^{4} = 5\cdot x^{4}\). Integrate, then plug in \(b\) and \(a\).

Find an antiderivative of \(5x^{4}\):

So we can take \(F(x)=x^{5}\).

By the Fundamental Theorem of Calculus,

Use the compound interest formula

with
\(P = 20000,\ r = 0.024,\ m = 4,\ t = 8.\)

Compute

then approximate \((1.006)^{32}\) and choose the closest option.

Use

with \(P=20000,\ r=0.024,\ m=4,\ t=8\).

Using a calculator,

so

The closest choice is $24,200.

Answer: B

Marginal cost is \(C'(x)\).
To get actual cost \(C(x)\), integrate \(C'(x)\), then use \(C(3)=80\) to find the constant.

1. Compute \[ C(x) = \int(6x+7)\,dx. \]
2. Use the condition “it costs $80 to produce 3 units” → \(C(3)=80\) to solve for the constant.

We have

Step 1: Integrate

Step 2: Use \(C(3)=80\)

So

Compare with options: this is option D.

Answer: D

Split the integral into three parts:

Use:

• power rule for \(2x\),
• log rule for \(\frac{3}{x}\),
• constant rule for \(7\).

Find an antiderivative \(F(x)\) such that:

then compute \(F(4)-F(1)\).

Compute:

Step 1: Find an antiderivative

Integrate term by term:

• \(\displaystyle \int 2x\,dx = x^{2}\),
• \(\displaystyle \int \frac{3}{x}\,dx = 3\ln|x|\),
• \(\displaystyle \int 7\,dx = 7x\).

So an antiderivative is

Step 2: Evaluate from 1 to 4

Compute \(F(4)\):

Compute \(F(1)\):

Now subtract:

Optionally approximate:

• \(\ln 4 \approx 1.386\),
• \(3\ln 4 \approx 4.158\),
• so total ≈ \(36 + 4.158 = 40.158\).

This is a product of two functions: \(x^{5}\) and \(\ln x\).
Use the product rule.

Let

Then

Let

Then

By the product rule:

You have an exponential of a function of \(x\).
Use the chain rule:

Let

Find \(u'(x)\), then multiply by \(e^{u(x)}\).

Let

Compute

Since \(j(x) = e^{u(x)}\), by the chain rule:

Average cost is

First write \(\overline{C}(x)\), then minimize it using derivative tests.

1. Compute \[ \overline{C}(x) = \frac{x^{3} + 7x + 1024}{x}. \] Simplify.
2. Compute \(\overline{C}'(x)\), set it equal to zero.
3. Solve for \(x\), and remember that \(x\) is in hundreds of units.

Average cost (per hundred units) is

Simplify: [ \begin{aligned} \overline{C}(x) &= \frac{x^{3}}{x}

• \frac{7x}{x}
• \frac{1024}{x} \ &= x^{2} + 7 + \frac{1024}{x}. \end{aligned} ]

So

To find where this is minimized, differentiate and set equal to zero:

Set \(\overline{C}'(x)=0\):

Since \(512 = 8^{3}\), we get

So the candidate for minimum average cost is \(x=8\) hundreds of units.

Now, instead of using the second derivative, we check values near \(x=8\):

Check a bit to the left, say \(x=7\):

Check a bit to the right, say \(x=9\):

Since \(\overline{C}(8)\) is smaller than nearby values \(\overline{C}(7)\) and \(\overline{C}(9)\),
we can conclude that \(\overline{C}(x)\) has an absolute minimum at \(x=8\) (for \(x>0\)).

Remember \(x\) is in hundreds of units, so the number of units is

You need to build a revenue function \(R(x)\) that includes:

• Ticket revenue = (ticket price)(number of people),
• Concession revenue = (average spending per person)(number of people).

Both depend on \(x\).

1. Ticket price: \(24 + x\).
2. Number of people: \(20{,}000 - 400x\).
3. Concession revenue: \(6(20{,}000 - 400x)\).
4. Total revenue: \[ R(x) = (24+x)(20{,}000 - 400x) + 6(20{,}000 - 400x). \]
5. Find derivative and solve to get max

Let \(x\) be the increase (in dollars) above $24.

Step 1: Express price and attendance in terms of \(x\)

• Ticket price:

  \[ p(x) = 24 + x. \]

• Attendance:

  \[ N(x) = 20{,}000 - 400x. \]

  (Every $1 increase loses 400 people.)

Step 2: Ticket revenue

Step 3: Concession revenue

Each person spends $6 on concessions, so

Step 4: Total revenue function

First expand \((24+x)(20{,}000 - 400x)\):

Next, the concession term:

Add them:

So

This is a downward-opening quadratic, so it has a maximum.

Step 5: Use calculus to find the maximum

Algebra note: We could find the maximum by using the vertex formula

\[ > x = -\frac{b}{2a} > \]

for a parabola.
But this is a calculus class, so we’ll find the maximum using the derivative.

Differentiate:

Set \(R'(x)=0\) to find critical points:

So the revenue is maximized when \(x = 10\).

Recall \(x\) is the increase above $24, so the ticket price is

That answers part (a).

Because \(R(x)\) is a downward-opening parabola (coefficient \(-400 < 0\)), this critical point is indeed a maximum.

Step 6: Find the maximum revenue

Compute \(R(10)\):

So the maximum total revenue (tickets + concessions) is

Answer summary

• (a) The team should charge $34 per ticket.
• (b) The maximum revenue is $640,000.