18.01 Solutions, Problem Set 1

• §2.2, #5

Use formula (8) to calculate f′(x_{0}) if f(x) is equal to

a) x^{2} − 4x − 5; Solution: Formula (8) states

f′(x_{0}) = lim_{Δx→0} \frac{f(x_{0} + Δx) − f(x_{0})}{Δx}.

The numerator in (8) is

f(x_{0} + Δx) − f(x_{0}) = (x_{0} + Δx)^{2} − 4(x_{0} + Δx) − 5 − (x_{0}^{2} − 4x_{0} − 5)
= x_{0}^{2} + 2x_{0}Δx + (Δx)^{2} − 4x_{0} − 4Δx − 5 − x_{0}^{2} + 4x_{0} + 5
= 2x_{0}Δx + (Δx)^{2} − 4Δx
= Δx(2x_{0} + Δx − 4).

Dividing by Δx gives 2x_{0} + Δx − 4, so

f′(x_{0}) = lim_{Δx→0}(2x_{0} + Δx − 4) = 2x_{0} − 4.

b) x^{2} − 2x + 1; Solution: The numerator is (x_{0} +Δx)^{2} −2(x_{0} +Δx)+1−(x_{0}^{2} −2x_{0} +1) = 2x_{0}Δx+(Δx)^{2} −2Δx,

so f′(x_{0}) = lim_{Δx→0}(2x_{0} + Δx − 2) = 2x_{0} − 2.

c) 2x^{2} + 1;

Solution: The numerator is

2(x_{0} + Δx)^{2} + 1 − (2x_{0}^{2} + 1) = 4x_{0}Δx + 2(Δx)^{2},

so f′(x_{0}) = lim_{Δx→0}(4x_{0} + 2Δx) = 4x_{0}.

d) x^{2} − 4;

Solution: The numerator is

(x_{0} + Δx)^{2} − 4 − (x_{0}^{2} − 4) = 2x_{0}Δx + (Δx)^{2},

so f′(x_{0}) = lim_{Δx→0}(2x_{0} + Δx) = 2x_{0}.

• §2.2, #8

Find the equations of the two lines through the point (3, 1) that are tangent to the curve y = x^{2} − 4. Hint: Draw the graph, let (a, a^{2} − 4) be the point of tangency, and find a.

Solution:

According to 5d), the tangent line at the the point (a, a^{2} − 4) has slope 2a. The equation of the tangent line at this point is therefore

y − (a^{2} − 4) = 2a(x − a).

For this line to pass through (3, 1), we must have

\begin{aligned}1 − (a^{2} − 4) & = & 2a(3 − a), \\ 5 − a^{2} & = & 6a − 2a^{2}, \\ a^{2} − 6a + 5 & = & 0, \\ (a − 5)(a − 1) & = & 0. \end{aligned}

The solutions are a = 1 and a = 5. The equations of the corresponding tangent lines are

\begin{matrix}y + 3 = 2(x − 1) & (or y = 2x − 5); \\ y − 21 = 10(x − 5) & (or y = 10x − 29). \end{matrix}

• Use the geometric interpretation of the derivative to explain why the derivative of a linear function is a constant function. (Use words and possibly pictures, but no formulas. Do not use pronouns.)

Solution: The derivative is the slope of the tangent line to the graph of the function. If the function is linear, then the graph of the function is a line, and the tangent line at any point is the same line. The slope of the tangent line is therefore the same at any point. Since the slope is constant, the derivative is constant.

• §2.3, #1

Use the three-step rule to show that if f(x) = ax^{2} + bx + c, then f′(x) = 2ax + b.

Solution: Step 1:

\begin{aligned}f(x + Δx) − f(x) & = & a(x + Δx)^{2} + b(x + Δx) + c − (ax^{2} + bx + c) \\ & = & 2axΔx + (Δx)^{2} + bΔx. \end{aligned}

Step 2:

\frac{f(x + Δx) − f(x)}{Δx} = 2ax + Δx + b.

Step 3:

f′(x) = lim_{Δx→0}(2ax + Δx + b) = 2ax + b.

• §2.3, #6

G(x) = 3x^{2} + 5x − 7; find G′(x).

Solution:

G′(x) = 6x + 5.

(Apply the rule with a = 3, b = 5 and c = −7.)

• §2.3, #18

Find all points on the curve y = f(x) = 6x − x^{2} at which the tangent is horizontal.

Solution: The rule in problem 1 gives f′(x) = 6 − 2x. The tangent line at a point is horizontal if f′(x) = 0. Solving the equation 6 − 2x = 0 gives x = 3, so the only point at which the tangent is horizontal is (3, 9) (since f(3) = 9).

• §2.3, #26 Use the three-step rule to calculate f′(x) if f(x) = x^{4}. Solution: Step 1: Using the binomial theorem,

f(x + Δx) − f(x) = (x + Δx)^{4} − x^{4}
= x^{4} + 4x^{3}Δx + 6x^{2}(Δx)^{2} + 4x(Δx)^{3} + (Δx)^{4} − x^{4}
= 4x^{3}Δx + 6x^{2}(Δx)^{2} + 4x(Δx)^{3} + (Δx)^{4}.

Step 2:

\frac{f(x + Δx) − f(x)}{Δx} = 4x^{3} + 6x^{2}Δx + 4x(Δx)^{2} + (Δx)^{3}.

Step 3:

f′(x) = lim_{Δx→0}(4x^{3} + 6x^{2}Δx + 4x(Δx)^{2} + (Δx)^{3}) = 4x^{3}.

• Sketch the graph of f(x) = x|x|. From the graph, predict the value of f′(0). Verify this prediction using the definition of the derivative. Solution: For x ≥ 0, we have f(x) = x^{2}, but if x < 0, then f(x) = −x^{2}.

The tangent line at (0, 0) appears to be horizontal, so we predict f′(0) = 0. According to the definition

f′(0) = lim_{Δx=0} \frac{f(0 + Δx) − f(0)}{Δx}.

The numerator is Δx|Δx|, so the quotient is |Δx| and

f′(0) = lim_{Δx=0} |Δx| = 0.

• §2.4, #2

The formula s = 1 − 6t − t^{2} describes the motion of a point along a horizontal line whose positive direction is to the right. Write down the velocity v = ds/dt by inspection. Also, find (a) the times when the velocity is zero, so that the point is momentarily at rest; and (b) the times when the point is moving to the right.

Solution: The velocity is

v = \frac{ds}{dt} = −6 − 2t.

a) The velocity is zero when −6 − 2t = 0, so −6 = 2t, so t = −3.

b) The point is moving the right when −6 − 2t > 0, so −6 > 2t, so t < −3.

• §2.4, #8

Two points start from the origin on the s-axis at time t = 0 and move along this axis in accordance with the formulas

s_{1} = t^{2} − 6t and s_{2} = 9t − 2t^{2},

where s_{1} and s_{2} are measured in feet and t in seconds.

a) When will the two points have the same speed?

b) What are the velocities of the two points at the times when they have the same position?

Solution: a) The velocities are

v_{1} = 2t − 6 and v_{2} = 9 − 4t.

The speeds |v_{1}| and |v_{2}| coincide when

2t − 6 = 9 − 4t (v_{1} = v_{2}, the points are moving in the direction)
or 2t − 6 = 4t − 9 (v_{1} = −v_{2}, the points are moving in opposite directions).

Solving the first equation gives 6t = 15, so t = 2.5 seconds. Solving the second equation gives 2t = 3, so t = 1.5 seconds.

b) The positions concide when

\begin{aligned}t^{2} − 6t & = 9t − 2t^{2}; \\ 3t^{2} − 15t & = 0; \\ 3t(t − 5) & = 0; \end{aligned}

so t = 0 and t = 5. The velocities when t = 0 are v_{1} = −6 ft/sec and v_{2} = 9 ft/sec. The velocities when t = 5 are v_{1} = 4 ft/sec and v_{2} = −11 ft/sec.

• p.81, #14

On a certain bicycle trip, the first half of the distance was covered at 30 mi/h and the second half at 20 mi/h. What was the average velocity? (It was not 25 mi/h.)

Solution: Let d be the total distance in miles. The first half of the trip, d/2 miles, was covered at 30 mi/h, so it took d/60 hours. The second half was covered at 20 mi/h, so it took d/40 hours. The total time taken in hours was therefore

\frac{d}{60} + \frac{d}{40} = \frac{2d}{120} + \frac{3d}{120} = \frac{5d}{120} = \frac{d}{24}.

The average velocity in mi/h was therefore

\frac{d}{d/24} = 24.

• §2.5 Some of the following limits exist, and others do not. Evaluate those that do.

– #2) lim_{x→2} \frac{10}{3 + x}.

Solution: lim_{x→2} \frac{10}{3 + x} = \frac{10}{5} = 2.

– #4) lim_{x→2} \frac{6}{2x − 4}

Solution: The denominator approaches zero, but the numerator does not, so the limit does not exist.

– #9) lim_{x→−3} (\frac{4x}{x + 3} + \frac{12}{x + 3}) .

Solution:

lim_{x→−3} (\frac{4x}{x + 3} + \frac{12}{x + 3}) = lim_{x→−3} \frac{4x + 12}{x + 3}
= lim_{x→−3} 4
= 4.

– #10) lim_{x→0.001} \frac{x}{|x|}

Solution: For x sufficiently near 0.001, x is positive, so \frac{x}{|x|} = 1.

Therefore

lim_{x→0.001} \frac{x}{|x|} = 1.

– #14) lim_{x→4} \frac{x − 4}{\sqrt{x} − 2}

Solution:

lim_{x→4} \frac{x − 4}{\sqrt{x} − 2} = lim_{x→4} \frac{(\sqrt{x} + 2)(\sqrt{x} − 2)}{\sqrt{x} − 2}
= lim_{x→4}(\sqrt{x} + 2)
= 4.

• §2.5, #25

Consider the limit

lim_{x→0^{+}} x^{x}.

a) Use a calculator to construct a table of values of x^{x} for x = 1, 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1, 0.05, 0.01, 0.005, 0.001. Use this evidence to form a conjecture about the value of the limit.

Solution: The following is a table of approximate values:

\begin{matrix}x & x^{x} \\ 1 & 1 \\ 0.9 & 0.9095 \\ 0.8 & 0.8365 \\ 0.7 & 0.7791 \\ 0.6 & 0.7360 \\ 0.5 & 0.7071 \\ 0.4 & 0.6931 \\ 0.3 & 0.6968 \\ 0.2 & 0.7248 \\ 0.1 & 0.7943 \\ 0.05 & 0.8609 \\ 0.01 & 0.9549 \\ 0.005 & 0.9738 \\ 0.001 & 0.9931 \end{matrix}

So we would conjecture that

lim_{x→0^{+}} x^{x} = 1.

b) Use the information in (a) to sketch a graph of y = x^{x} for 0 < x < 1. Estimate the location of the lowest point.

Solution:

The lowest point is near x = 0.4. (Later we’ll see that the lowest point is at x = e^{−1}.)

• Give an example of a function which is continuous at every point, and is differentiable at every point except x = 1 and x = −1.

Solution: Let f(x) = |x^{2} − 1|.

¿From its graph, we see that it is continuous at every point, and differentiable at every point except x = 1 and x = −1.