AP Unit 1 Solutions

1. Compute s(2)=10 and s(2+h)=10+7h+h². Reason: substitution and expansion keep the same position model.
2. Divide the change by h: [s(2+h)−s(2)]/h=7+h for h≠0. Reason: average rate is change per elapsed time.
3. Take h→0 to obtain 7 m/s. Reason: the instantaneous rate is the limiting average rate, not a zero-length quotient.

Answer: Average rate 7+h m/s; instantaneous rate 7 m/s.

1. Use t→5⁻ to indicate approach from smaller time inputs. Reason: the superscript minus specifies the side.
2. Write lim(t→5⁻)C(t)=8 mg/L. Reason: a limit records the approached output and its units.
3. No value for C(5) is determined. Reason: one-sided nearby behavior does not specify the actual target value.

Answer: lim(t→5⁻)C(t)=8 mg/L; C(5) remains unspecified.

1. Compare the two one-sided limits: both are 2. Reason: equality is required for a finite two-sided limit.
2. Conclude lim(x→4)f(x)=2. Reason: both sides approach the same output.
3. The value −1 occurs only at x=4. Reason: the definition of the limit excludes that one input.

Answer: The limit is 2 although f(4)=−1.

1. Trace toward the open point from each side: both nearby heights approach 3. Reason: the graph follows y=x+1 away from 2.
2. The two-sided limit is 3. Reason: the one-sided limits agree.
3. Read the filled point to get f(2)=5. Reason: a filled point represents the included function value.

Answer: Left limit 3; right limit 3; two-sided limit 3; f(2)=5.

1. Choose xₙ=1/(π/2+2πn) and yₙ=1/(3π/2+2πn). Both approach 0 from the right. Reason: the denominators grow without bound.
2. The corresponding outputs are 1 and −1 for every n. Reason: these angles are sine maxima and minima.
3. The limit does not exist; the plot undersamples the oscillations. Reason: arbitrarily nearby inputs retain different output values.

Answer: No limit exists; graph scale and sampling can hide persistent oscillation.

1. Read the left entries as x moves toward 1: 2.71 then 2.9701 approach 3. Reason: the inputs approach from below.
2. Read the right entries toward 1: 3.31 then 3.0301 also approach 3. Reason: a two-sided estimate needs both sides.
3. Estimate 3, but do not claim proof from only these samples. Reason: no finite table controls every sufficiently nearby input.

Answer: Estimated limit 3; the table alone does not prove it.

1. Numerator approaches 5 and denominator approaches 5. Reason: polynomials are continuous.
2. The denominator limit is nonzero. Reason: this is the hypothesis for the quotient limit rule.
3. Divide 5 by 5 to obtain 1. Reason: the quotient rule for limits now applies.

Answer: 1; the limiting denominator is 5≠0.

1. Use x² on the left: lim(x→1⁻)f(x)=1. Reason: the left-hand inputs use the first branch.
2. Use 2x+1 on the right: lim(x→1⁺)f(x)=3. Reason: the right-hand inputs use the second branch.
3. The two-sided limit does not exist. Reason: 1≠3.

Answer: Left 1; right 3; two-sided limit does not exist.

1. Factor x²+x−2=(x+2)(x−1). Reason: factoring reveals the zero-producing factor.
2. For x≠−2, cancel x+2 to obtain x−1. Reason: nearby expressions may be replaced by equal expressions.
3. Take x→−2 to get −3. Reason: the simplified linear function is continuous.

Answer: −3.

1. Multiply by the conjugate √(9+x)+3 in numerator and denominator. Reason: a difference of squares removes the radical difference.
2. For x≠0 the result is 1/(√(9+x)+3). Reason: the numerator simplifies to x and cancels.
3. Substitute x=0 in the continuous simplified expression to get 1/6. Reason: its denominator approaches 6≠0.

Answer: 1/6.

1. Rewrite as (4/3)[sin(4x)/(4x)]. Reason: the standard sine limit needs the same argument in numerator and denominator.
2. As x→0, 4x→0 and sin(4x)/(4x)→1. Reason: the standard radian sine limit applies.
3. Multiply by 4/3. Reason: constant multiples pass through finite limits.

Answer: 4/3.

1. (a) Choose factoring, then cancel x−2 to get x+2→4. Reason: a difference of squares resolves 0/0.
2. (b) Choose a conjugate, producing 1/(√(1+x)+1)→1/2. Reason: the radical difference becomes a factor x.
3. (c) Choose sign and magnitude analysis to get +∞. Reason: a positive denominator shrinking to 0 makes the quotient unbounded.

Answer: (a) Factoring: 4. (b) Conjugate: 1/2. (c) One-sided sign analysis: +∞.

1. Use −1≤cos(1/x)≤1 for x≠0. Reason: cosine is bounded.
2. Multiply by nonnegative x²: −x²≤x²cos(1/x)≤x². Reason: multiplication by a nonnegative number preserves inequality order.
3. Both bounding limits are 0, so the middle limit is 0. Reason: the squeeze theorem applies near 0.

Answer: 0, justified by bounds −x² and x².

1. The lower bound approaches 5 as x→1. Reason: |x−1|→0.
2. The upper bound also approaches 5; conclude the middle limit is 5. Reason: the stated punctured neighborhood satisfies the squeeze hypotheses.
3. No value of g(1) is forced. Reason: the inequality deliberately excludes the target input.

Answer: The limit is 5; g(1) need not be 5.

1. Factor and cancel to get H(x)=x+3 for x≠3. Reason: this expression gives the same nearby outputs.
2. Compute H(2.99)=5.99 and H(3.01)=6.01; the graph is y=x+3 with a hole at (3,6). Reason: the missing input does not alter the rest of the line.
3. Write lim(x→3)H(x)=6: the output approaches 6 as the input approaches 3. Reason: both sides approach the same height.

Answer: Table 5.99 and 6.01; hole at (3,6); limit 6; H(3) remains undefined.

1. (a) The nearby quotient equals x+2 and tends to 4. Reason: the common factor cancels, so this is removable and f(2)=4 repairs it.
2. (b) The side limits are −1 and 1. Reason: they disagree, so this is a jump and one value cannot repair it.
3. (c) Both sides are unbounded positive. Reason: a positive square denominator shrinks to 0, so this is infinite and cannot be repaired with a finite value.

Answer: (a) Removable, repair with 4. (b) Jump, no repair. (c) Infinite, no finite repair.

1. f(2)=4 exists. Reason: the target rule defines it.
2. The two-sided limit is 5. Reason: nearby values follow the continuous line 2x+1.
3. Since 5≠4, continuity fails. Reason: the third condition requires equality of value and limit.

Answer: Not continuous; the value-limit equality fails.

1. Read g(1)=1 from x². Reason: that branch includes the target.
2. The left limit is 1²=1 and the right limit is 2(1)−1=1. Reason: each branch is continuous on its side.
3. Thus the finite limit exists and equals g(1)=1. Reason: all three continuity conditions hold.

Answer: Yes, g is continuous at 1.

1. Factor x²−4=(x−2)(x+2); the original denominator excludes x=−2 and x=2. Reason: the original expression must be defined.
2. A rational function is continuous wherever its denominator is nonzero. Reason: the quotient of continuous functions is continuous on that domain.
3. State (−∞,−2), (−2,2), (2,∞). Cancellation does not add −2. Reason: an equivalent nearby rule does not redefine the original function.

Answer: (−∞,−2), (−2,2), (2,∞); cancellation does not change the original domain.

1. Require x−2>0, so x>2. Reason: the real logarithm needs a positive argument.
2. ln is continuous for positive arguments and x−2 is continuous. Reason: a valid composition of continuous functions is continuous.
3. The interval is (2,∞), excluding 2. Reason: ln 0 is undefined.

Answer: (2,∞).

1. Factor x³−8=(x−2)(x²+2x+4). Reason: a difference of cubes reveals the common factor.
2. The nearby function becomes x²+2x+4 and its limit at 2 is 12. Reason: cancellation is valid for x≠2 and the remaining polynomial is continuous.
3. Set f(2)=12. Reason: continuity requires the actual value to equal that finite limit.

Answer: f(2)=12.

1. The target value and right limit equal 5. Reason: the second branch includes x=1 and is continuous.
2. The left limit is a+2. Set a+2=5. Reason: both sides and the value must agree.
3. Solve a=3, then check both branches give limiting value 5. Reason: substitution verifies the parameter repair.

Answer: a=3.

1. On the left, x−3 is negative and near 0, so −2/(x−3) is unbounded positive. Reason: dividing two negative quantities gives a positive quotient.
2. On the right, x−3 is positive and near 0, so the quotient is unbounded negative. Reason: the fixed numerator is negative.
3. State left limit +∞, right limit −∞ and asymptote x=3. Reason: unbounded one-sided behavior defines a vertical asymptote.

Answer: Left +∞; right −∞; vertical asymptote x=3.

1. Near 2 the numerator is positive and approaches 3. Reason: the linear numerator is continuous.
2. The squared denominator is positive on both sides and approaches 0. Reason: a nonzero square is positive.
3. The quotient grows without bound on both sides, so lim(x→2)W(x)=+∞ in infinite-limit notation. Reason: the signs agree; this describes unbounded growth, not a finite real limit.

Answer: Unbounded positive on both sides; +∞ is not a finite real limit.

1. Divide top and bottom by x² for x≠0. Reason: highest powers control end behavior.
2. The quotient becomes (4−3/x+1/x²)/(2+7/x²). Reason: division preserves the expression.
3. At either end the inverse powers vanish, giving 4/2=2. Reason: the limiting denominator is nonzero.

Answer: Both limits are 2; horizontal asymptote y=2.

1. Write the quotient as (|x|/x)√(4+1/x²). Reason: √(x²)=|x|.
2. For positive x, |x|/x=1 and the radical tends to 2. Reason: 1/x²→0.
3. For negative x, |x|/x=−1, so the limit is −2. Reason: the sign factor differs at the two ends.

Answer: Positive end: 2; negative end: −2.

1. Compute approximate ratios 0.541341, 0.168449, 0.004540 and 0.000000824. Reason: division compares the sizes in a common scale.
2. The stated limit 0 means x² becomes negligible compared with eˣ. Reason: a vanishing ratio describes relative growth.
3. The finite table supports the trend but does not prove all far-out behavior. Reason: four inputs cannot establish a statement about every sufficiently large input.

Answer: eˣ dominates x² in the ratio sense; four samples are evidence, not a proof.

1. Let f(x)=x³−2x−2; it is continuous on [1,2]. Reason: it is a polynomial.
2. f(1)=−3 and f(2)=2, so 0 lies strictly between them. Reason: the desired output is bracketed.
3. IVT guarantees a c in (1,2) with f(c)=0. Reason: continuity and the intermediate output condition hold.

Answer: At least one root lies in (1,2), by IVT.

1. 19 lies strictly between 14 and 22. Reason: the endpoint temperatures bracket the target.
2. Continuity on [0,8] lets IVT guarantee at least one t in (0,8) with T(t)=19. Reason: a continuous function cannot skip that output.
3. Uniqueness is not guaranteed. Reason: temperature could rise and fall and cross 19 more than once.

Answer: Yes, at least once; exactly once is not guaranteed.

1. The left limit at 0 is −1 and the right limit is 1. Reason: each side is a constant horizontal graph.
2. The function is not continuous on [−2,2], so IVT does not apply. Reason: continuity on the complete interval is a required hypothesis.
3. Its only output values are −1 and 1, so no zero exists. Reason: the explicit rule never outputs 0.

Answer: IVT fails because of the jump; the function has no zero.