MATH 116 Test 3 Practice (Fall 2024)

When differentiating \(\ln(5x+7)\), what role does the inside function play compared with differentiating \(\ln x\)?

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

How can you rewrite both sides so they have the same exponential base, and what equation does that let you solve?

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

If you know a rate function \(f'(t)\) and an initial value \(f(0)\), how do you recover the original quantity in a later year?

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

What two ingredients do you always need for a tangent-line equation at \(x=0\)?

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

If the effective rate \(E\) is given for continuous compounding, which equation connects \(E\) and the nominal rate \(r\)?

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

If \(f'(x)=10x+2\), what family of functions could produce that derivative?

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

Which integration rule matches each term in this integrand: \(e^{4x}\) and \(x^{-3}\)?

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):

Since \(f(x)=\dfrac{x}{3+\ln x}\), what derivative rule is the best starting point?

Let

Then

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

Write

Let

Compute derivatives:

By the quotient rule,

To apply the Fundamental Theorem of Calculus here, what antiderivative do you need first?

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,

Which quantities in the compound-interest formula represent principal, annual rate, compounding frequency, and time in this situation?

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

How do you move from marginal cost \(C'(x)\) to total cost \(C(x)\), and where is the given value \(C(3)=80\) used?

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

For \(\int_1^4\left(2x+\frac{3}{x}+7\right)dx\), why is it useful to split the integrand into separate pieces before integrating?

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 ≈ \(36 + 4.158 = 40.158\).

For \(h(x)=x^5\ln x\), which derivative rule combines these two factors?

Let

Then

Let

Then

By the product rule:

For \(j(x)=e^{\sqrt{x}+5x^2}\), what is the outer function and what is the inner function of \(x\)?

Let

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

Let

Compute

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

What expression represents average cost, and what condition identifies where that average cost can be minimized?

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

Which quantities change with \(x\), and how should ticket revenue and concession revenue be combined into one total revenue model?

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.