AP Unit 2 Solutions

1. Compute s(1)=−1 and s(4)=8; average rate=(8−(−1))/(4−1)=3 m/s. Reason: a secant slope compares two endpoints.
2. At 1, [s(1+h)−s(1)]/h=h for h≠0. Reason: s(1+h)=−1+h².
3. Let h→0 to obtain 0 m/s. Reason: the instantaneous rate is the limiting secant slope.

Answer: Average 3 m/s; instantaneous at 1 is 0 m/s.

1. Compute f(−2)=5, so the numerator is x²−4. Reason: the difference quotient compares nearby values with the base value.
2. Factor and cancel: (x²−4)/(x+2)=x−2 for x≠−2. Reason: equal nearby expressions have the same limit.
3. Take x→−2 to obtain −4. Reason: the simplified line is continuous.

Answer: f′(−2)=−4.

1. Subtract B(3) and divide by h to obtain 2+0.4h for h≠0. Reason: this is the average-rate quotient.
2. Take h→0 to get 2. Reason: the finite limit defines the derivative.
3. Attach units thousands of bacteria per unit time. Reason: derivative units are output units divided by the input units.

Answer: B′(3)=2 thousand bacteria per unit time.

1. Compute f(x+h)−f(x)=2xh+h²−3h. Reason: expansion isolates the output change.
2. For h≠0 the difference quotient is 2x+h−3. Reason: every numerator term has a factor h.
3. Let h→0 to obtain f′(x)=2x−3. Reason: the finite limit exists for every real x.

Answer: f′(x)=2x−3.

1. The tangent point is (−1,3). Reason: substitute into f.
2. f′(x)=2x, so f′(−1)=−2. Reason: the derivative gives tangent slope.
3. Use point-slope form y−3=−2(x+1), or y=−2x+1. Reason: a line is determined by its slope and point.

Answer: y=−2x+1.

1. Compute tangent slope (1−5)/(4−2)=−2. Reason: both supplied points lie on the tangent line.
2. Write F′(2)=−2 and tangent y−5=−2(x−2). Reason: the derivative equals the tangent slope.
3. F(4) need not be 1. Reason: a point on a tangent away from tangency need not lie on the curve.

Answer: F′(2)=−2; tangent y=−2x+9; F(4) is not determined.

1. D(6)=1.8 gives the depth at minute 6. Reason: D is the original output function.
2. D′(6)=−0.12 means depth is decreasing at an instantaneous rate of 0.12 m/min at that time. Reason: the derivative is a signed rate.
3. Write y−1.8=−0.12(t−6). Reason: use the depth and rate at the same input.

Answer: Depth 1.8 m; instantaneous decrease 0.12 m/min; tangent y−1.8=−0.12(t−6).

1. Use the points on either side of 2. Reason: a centered difference samples both sides.
2. Compute (6.84−5.24)/(2.2−1.8)=4. Reason: this is the nearby secant slope.
3. Report s′(2)≈4 m/s. Reason: measured data give an estimate of position change per second.

Answer: Approximately 4 m/s.

1. Use the two points on the tangent: slope=(6−(−2))/(5−1). Reason: a tangent line has the same slope at all its points.
2. Compute 8/4=2. Reason: slope is rise divided by run.
3. Conclude f′(3)≈2; distant curve points give a secant average instead. Reason: the derivative describes the local tangent rate.

Answer: f′(3)≈2.

1. At h=0.01, [f(1.01)−f(0.99)]/0.02≈−0.735734. Reason: the centered quotient estimates the local slope.
2. At h=0.001, [f(1.001)−f(0.999)]/0.002≈−0.735759. Reason: checking a second step size tests the stability of the estimate.
3. Report about −0.735759; very small h can introduce cancellation or rounding error. Reason: technology approximates a limiting process with finite precision.

Answer: f′(1)≈−0.735759.

1. f(2)=0 and |x−2|→0 as x→2, so it is continuous. Reason: value and limit agree.
2. The base-point quotient is |h|/h, equal to −1 for h<0 and 1 for h>0. Reason: the absolute value depends on the sign.
3. No derivative exists at 2. Reason: the one-sided limiting slopes disagree.

Answer: Continuous at 2; not differentiable there.

1. The nearby graph approaches height 1 from both sides. Reason: x² is continuous.
2. The filled point gives f(1)=7, so the graph is discontinuous at 1. Reason: the value differs from the limit.
3. It is not differentiable at 1. Reason: differentiability implies continuity.

Answer: Not differentiable at 1, because it is discontinuous there.

1. The function value and limit at 0 are both 0. Reason: the real cube-root function is continuous.
2. For h≠0, [f(h)−f(0)]/h=h^(−2/3), which grows without bound positively as h→0. Reason: h^(2/3)>0 and shrinks to 0.
3. There is a vertical tangent and no finite derivative at 0. Reason: a derivative requires a finite limiting slope.

Answer: Continuous; vertical tangent; no finite derivative at 0.

1. Apply the power rule: f′(x)=7x⁶. Reason: multiply by 7 and reduce the exponent by 1.
2. Evaluate at −1: 7(−1)⁶. Reason: the requested slope uses the derivative function.
3. The result is 7. Reason: an even power of −1 is 1.

Answer: f′(x)=7x⁶; f′(−1)=7.

1. Write f(x)=x^(−4), x≠0. Reason: power notation exposes the negative exponent.
2. Differentiate to −4x^(−5). Reason: the power rule also applies to negative powers on their domains.
3. Rewrite as −4/x⁵ and retain x≠0. Reason: differentiation does not fill a missing input.

Answer: f′(x)=−4/x⁵ for x≠0.

1. For x>0, f′(x)=(3/2)√x. Reason: subtract 1 from the exponent 3/2.
2. At 0 from the right, [h^(3/2)−0]/h=√h→0. Reason: allowed h must be positive in the real domain.
3. This gives a right endpoint rate of 0, rather than a two-sided derivative on a real open neighborhood. Reason: the real function is undefined for negative x.

Answer: For x>0, f′(x)=(3/2)√x; right endpoint rate at 0 is 0, with no two-sided real-neighborhood derivative asserted.

1. Differentiate 5x⁴ to 20x³. Reason: constant multiple and power rules.
2. Differentiate −3x² and 2x to −6x and 2. Reason: linearity preserves coefficients and signs.
3. Differentiate −11 to 0 and combine terms. Reason: a constant has no change.

Answer: P′(x)=20x³−6x+2.

1. A′(x)=6x²−4 and B′(x)=2x. Reason: polynomial derivatives use power and linearity rules.
2. (3A−2B)′=3(6x²−4)−2(2x). Reason: derivatives pass through constant multiples and differences.
3. At 1 the result is 3·2−2·2=2. Reason: evaluate after differentiating.

Answer: (3A−2B)′(1)=2.

1. The trig terms give 3cos x+2sin x. Reason: cosine differentiation introduces a minus sign that cancels the coefficient sign.
2. The exponential and logarithm terms give eˣ+1/x. Reason: use the familiar derivative rules.
3. Combine and require x>0. Reason: the logarithm restricts the original domain.

Answer: f′(x)=3cos x+2sin x+eˣ+1/x, x>0.

1. Recognize the derivative definition for f(x)=sin x at a=π/6. Reason: the numerator is f(a+h)−f(a).
2. Use f′(x)=cos x in radians. Reason: the known rule evaluates the limiting quotient.
3. Compute cos(π/6)=√3/2. Reason: this is a familiar unit-circle value.

Answer: √3/2.

1. The function is f(x)=ln x and the base input is 2. Reason: the expression matches [f(x)−f(2)]/(x−2).
2. Its limit is f′(2). Reason: this is the x-form derivative definition.
3. Since f′(x)=1/x, the value is 1/2. Reason: 2 lies in the logarithm domain.

Answer: Function ln x, base 2, limit 1/2.

1. The tangent point is (0,1). Reason: e⁰=1.
2. The derivative is eˣ, so the slope at 0 is 1. Reason: the exponential rate equals its value.
3. Use y−1=1(x−0). Reason: point-slope form uses the tangent point and rate.

Answer: y=x+1.

1. Identify derivatives 2x and −sin x for the two factors. Reason: both factors are differentiable.
2. Apply the product rule: 2x cos x+x²(−sin x). Reason: each factor contributes a rate term.
3. Simplify signs to 2x cos x−x²sin x. Reason: combining signs preserves the formula.

Answer: f′(x)=2x cos x−x²sin x.

1. Write (fg)′(2)=f′(2)g(2)+f(2)g′(2). Reason: the product rule applies to both differentiable factors.
2. Substitute 3·4+(−1)(−2). Reason: all values use input 2.
3. Add 12+2=14. Reason: the negative product contributes positively.

Answer: 14.

1. A(t)=L(t)W(t). Reason: rectangular area is the product of its dimensions.
2. A′=L′W+LW′=−0.4·2+5·0.6. Reason: both dimensions change.
3. Compute A′=2.2 cm²/s. Reason: area-change units come from length-rate times length.

Answer: Area is increasing at 2.2 cm²/s.

1. Use numerator derivative 2x and denominator derivative 1. Reason: both are polynomials.
2. Form [2x(x+1)−(x²+2)]/(x+1)². Reason: the quotient rule keeps numerator order and squares the denominator.
3. Simplify to (x²+2x−2)/(x+1)² with x≠−1. Reason: the original denominator still restricts the domain.

Answer: f′(x)=(x²+2x−2)/(x+1)², x≠−1.

1. Check g(1)=3≠0. Reason: a quotient must be defined at the requested input.
2. Use [f′(1)g(1)−f(1)g′(1)]/g(1)²=[−2·3−6·4]/9. Reason: the quotient rule applies.
3. Compute −30/9=−10/3. Reason: exact reduction avoids rounding.

Answer: −10/3.

1. For x≠0, rewrite R=x+1/x. Reason: divide each numerator term by x.
2. Differentiate to 1−1/x². Reason: use the sum and negative-power rules.
3. The quotient rule gives [2x·x−(x²+1)]/x²=(x²−1)/x², the same result. Reason: equivalent domain-preserving methods should agree.

Answer: R′(x)=1−1/x², x≠0.

1. Use (tan x)′=sec²x and (cot x)′=−csc²x. Reason: these follow from trig identities and the quotient rule.
2. Thus f′=sec²x+csc²x. Reason: subtracting cotangent cancels its derivative minus sign.
3. At π/4 each squared reciprocal is 2, so f′=4. Reason: sine and cosine both equal √2/2.

Answer: f′(x)=sec²x+csc²x on the common domain; f′(π/4)=4.

1. Differentiate to sec x tan x−csc x cot x. Reason: the reciprocal-trig rules have different signs.
2. At π/6, sec·tan=(2/√3)(1/√3)=2/3 and csc·cot=2√3. Reason: use exact unit-circle values.
3. The value is 2/3−2√3; exclude sin x=0 or cos x=0. Reason: both original terms must be defined.

Answer: f′(x)=sec x tan x−csc x cot x; f′(π/6)=2/3−2√3; sin x and cos x must both be nonzero.