Chapter 3 Solutions

Authored worked explanation for the original source problem.

1. At x=−1, nearby inputs lie in the branch $f\left(x\right)=x^2+1$ because −1<1.

   Why: The remote branch at x>1 does not affect this local slope.

2. Differentiate to get $f\prime \left(x\right)=2x$ and substitute −1 to get −2.

   Why: The polynomial power rule applies in this neighborhood.

Reviewed answer: $-2$

Takeaway: A piecewise derivative uses the branch near the point.

Authored worked explanation for the original source problem.

1. Divide each numerator term by x: $f\left(x\right)=2x^2+4x+3+x^{-2}$.

   Why: This avoids introducing a quotient rule before the next chapter.

2. Differentiate term by term to get $4x+4-2x^{-3}$.

   Why: Use the power rule and the zero derivative of the constant.

Reviewed answer: $4x+4-2x^{-3}$, for x≠0.

Takeaway: Algebra can turn a quotient into a sum of powers.

Authored worked explanation for the original source problem.

Source review note: The source groups this under “using only the power rule,” but e^x requires its special derivative. The source derivative itself is consistent.

1. Rewrite $4/x=4x^{-1}$ and $\sqrt{x}=x^{1/2}$.

   Why: Fractional and negative powers fit the power rule.

2. Differentiate the three terms to get the answer.

   Why: The exponential term uses d(e^x)/dx=e^x, so the prompt's “only the power rule” instruction cannot literally cover this term.

Reviewed answer: $3e^x-4x^{-2}+(1/2)x^{-1/2}$, for x>0.

Takeaway: Choose the derivative rule appropriate to each term.

Authored worked explanation for the original source problem.

Source review note: The source gives sec x tan x−sin x. The printed function simplifies to tan x+cos x, whose derivative is sec²x−sin x.

1. Use $\tan x·\sin x={\sin }^{2}x/\cos x=\sec x-\cos x$.

   Why: Rewrite sin² as 1−cos² to avoid a product rule.

2. Then $f\left(x\right)=\sec x+\tan x-(\sec x-\cos x)=\tan x+\cos x$.

   Why: The two secant terms cancel.

3. Differentiate: $f\prime \left(x\right)={\sec }^{2}x-\sin x$.

   Why: Use the known tangent and cosine derivatives.

Reviewed answer: ${\sec }^{2}x-\sin x$, where cos x≠0.

Takeaway: Simplification can reveal cancellations before differentiation.

Authored worked explanation for the original source problem.

Source review note: The source derivative also adds 2cos x, but the printed cos x cot x+sin x simplifies to csc x. The first radical equals |sin x|, so the reviewed derivative additionally needs sign cases.

1. $\sqrt{1-{\cos }^{2}x}=\left|\sin x\right|$, so the first term is $\cos x/\left|\sin x\right|$.

   Why: The nonnegative square root is an absolute value, not automatically sin x.

2. The remaining terms simplify: $\cos x\cot x+\sin x={\cos }^{2}x/\sin x+\sin x=\csc x$.

   Why: Use cos²+sin²=1, with sin x≠0.

3. On sin x>0, $f=\cot x+\csc x$; on sin x<0, $f=-\cot x+\csc x$.

   Why: Treat each sign interval separately.

4. Hence the derivatives are respectively $-{\csc }^{2}x-\csc x\cot x$ and ${\csc }^{2}x-\csc x\cot x$.

   Why: The sign is constant on each interval; differentiate the simplified trig functions.

Reviewed answer: For sin x>0: $-{\csc }^{2}x-\csc x\cot x$. For sin x<0: ${\csc }^{2}x-\csc x\cot x$.

Takeaway: A square root of a square must keep the absolute value.

Authored worked explanation for the original source problem.

1. $f\prime \left(x\right)=3(x-1{)}^2$ and $x_{n+1}=x_n-f\left(x_n\right)/f\prime \left(x_n\right)$.

   Why: Newton's next guess is the tangent line's x-intercept.

2. At 0, f=3 and f′=3, so $x_1=-1$; at −1, f=−4 and f′=12, so $x_2=-2/3$.

   Why: Recompute both height and slope at each new guess.

3. At −2/3, $f=-17/27$ and $f\prime =25/3$, giving $x_3=-2/3+17/225=-133/225$.

   Why: The derivative is nonzero at all three iteration inputs.

Reviewed answer: $x_1=-1$, $x_2=-2/3$, $x_3=-133/225$.

Takeaway: Newton uses the new guess for both function and derivative values.

Authored worked explanation for the original source problem.

1. $x_1=e-\ln \left(e\right)/(1/e)=e-e=0$.

   Why: The first iteration has a valid positive input.

2. The next step needs ln(0), which is undefined.

   Why: An iteration must stay in the function's domain to continue.

Reviewed answer: $x_1=0$; the second iteration is undefined.

Takeaway: A legal initial guess can produce an illegal next guess.

Authored worked explanation for the original source problem.

Source review note: The source repeats the sine/cosine answer from problem 9. The printed polynomial/reciprocal function gives 12x+8+2/x³.

1. The first derivative is $6x^2+8x+3-x^{-2}$.

   Why: Apply the power rule to every term.

2. Differentiate once more: $12x+8+2x^{-3}$.

   Why: The first derivative's constant disappears and the reciprocal term changes sign again.

Reviewed answer: $12x+8+2x^{-3}$, for x≠0.

Takeaway: A second derivative is the derivative of the whole first derivative.

Authored worked explanation for the original source problem.

1. $f\prime \left(x\right)=\cos x+\sin x$.

   Why: The minus sign on −cos cancels the derivative's minus sign.

2. $f\prime \prime \left(x\right)=-\sin x+\cos x$.

   Why: Differentiate both first-derivative terms.

Reviewed answer: $-\sin x+\cos x$

Takeaway: Track the sign on each trip around the trig derivative cycle.

Authored worked explanation for the original source problem.

Source review note: The source gives 3(e^x+1/√x), but (4√x)³=64x^(3/2), yielding 48/√x in the second derivative. At x=0 the derivatives need separate endpoint treatment.

1. $(x^2-1)/(x+1)=x-1$ on its original domain, and $(4\sqrt{x}{)}^3=64x^{3/2}$.

   Why: The coefficient 4 must also be cubed.

2. For x>0, $f\prime =3e^x+1+96\sqrt{x}$.

   Why: The polynomial simplifies before differentiation.

3. Then $f\prime \prime =3e^x+48/\sqrt{x}$.

   Why: Differentiate the square root using its fractional power.

Reviewed answer: $3e^x+48/\sqrt{x}$ for x>0.

Takeaway: A power applies to the entire parenthesized product.

Authored solution to an added practice problem.

1. Differentiate $5x^4$, −3x², 2x, and −7 separately.

   Why: A derivative distributes across sums.

2. The terms give $20x^3$, −6x, 2, and 0.

   Why: Multiply each power by its exponent and reduce the exponent by one.

Reviewed answer: $20x^3-6x+2$

Takeaway: A constant's derivative vanishes.

Authored solution to an added practice problem.

1. Rewrite the function as $x^{-1/2}+x^{3/2}$.

   Why: Square roots and reciprocal roots are powers.

2. Apply the power rule to both exponents.

   Why: Negative and fractional exponents follow the same algebraic rule on the positive domain.

Reviewed answer: $-(1/2)x^{-3/2}+(3/2)x^{1/2}$, for x>0.

Takeaway: Choose a form suited to the rule you know.

Authored solution to an added practice problem.

1. The derivative of 2sin x is 2cos x.

   Why: Constant multiples pass through differentiation.

2. The derivative of −3cos x is 3sin x.

   Why: The two negative signs cancel.

Reviewed answer: $2\cos x+3\sin x$

Takeaway: A quick sign check prevents the common cosine error.

Authored solution to an added practice problem.

1. Differentiate 4e^x to 4e^x.

   Why: The natural exponential reproduces itself.

2. Differentiate −tan x to −sec²x.

   Why: Use the tangent derivative on intervals where cosine is nonzero.

Reviewed answer: $4e^x-{\sec }^{2}x$, with $x\ne \pi /2+k\pi$ for any integer k.

Takeaway: The derivative exists only on the original function's domain.

Authored solution to an added practice problem.

1. For $f\left(x\right)=x^3$, $f\left(2\right)=8$ and $f\prime \left(2\right)=3·2^2=12$.

   Why: A tangent line needs a point and an instantaneous slope.

2. Use point-slope form $y-8=12(x-2)$, then simplify.

   Why: The line must pass through the same point as the curve.

Reviewed answer: $y=12x-16$.

Takeaway: The derivative supplies the slope, not the line's intercept.

Authored solution to an added practice problem.

1. Continuity at 2 requires $4=2a+b$.

   Why: A derivative cannot exist across a jump.

2. The left derivative is $2x$, so its value at 2 is 4; the right derivative is a.

   Why: One-sided slopes must match as well.

3. Set a=4, then the continuity equation gives b=−4.

   Why: Matching slope alone does not determine the vertical placement.

Reviewed answer: $a=4$ and $b=-4$.

Takeaway: A smooth join needs both equal heights and equal slopes.

Authored solution to an added practice problem.

1. For x<3, $|x-3|=3-x$; for x>3 it is x−3.

   Why: Split according to the sign of the inside expression.

2. The branch slopes are −1 and 1, so the one-sided slopes disagree at 3.

   Why: A corner prevents a two-sided derivative.

Reviewed answer: Derivative −1 for x<3, +1 for x>3; undefined at 3.

Takeaway: Translate the corner without changing its two slopes.

Authored solution to an added practice problem.

1. For $f\left(x\right)=x^2-3$, use $f\prime =2x$ and $x_{n+1}=x_n-({x_n}^2-3)/\left(2x_n\right)$.

   Why: The tangent intercept refines the estimate of √3.

2. From 2, $x_1=2-1/4=7/4$.

   Why: The initial height is 1 and slope is 4.

3. At 7/4, the height is 1/16 and slope 7/2, so $x_2=7/4-1/56=97/56$.

   Why: Update the slope along with the guess.

Reviewed answer: $x_1=7/4$, $x_2=97/56\approx 1.73214$.

Takeaway: Keep fractions to avoid accumulated rounding.

Authored solution to an added practice problem.

1. $f\prime \left(x\right)=3x^2$, giving zero slope at the proposed initial guess.

   Why: The tangent line is horizontal there.

2. $x_1=0-(-1)/0$ cannot be computed.

   Why: A horizontal tangent at a nonzero height has no x-intercept.

Reviewed answer: The first step is undefined because $f\prime \left(0\right)=0$ while $f\left(0\right)=-1$.

Takeaway: Check the denominator before starting Newton's formula.

Authored solution to an added practice problem.

1. $f\prime =6x^2-6x$, then $f\prime \prime =12x-6$.

   Why: Differentiate in two separate passes.

2. The second derivative is negative below 1/2 and positive above it.

   Why: Its sign describes how the slope changes.

Reviewed answer: $f\prime \prime \left(x\right)=12x-6$; concave down for x<1/2 and up for x>1/2.

Takeaway: Concavity concerns changing slopes, not just increasing heights.

Authored solution to an added practice problem.

1. Differentiate position $s\left(t\right)=t^3-6t^2+9t$ to get velocity.

   Why: A position rate has units of meters per second.

2. Differentiate velocity to get acceleration, then substitute t=1.

   Why: Acceleration measures the rate of change of velocity, even at an instant of rest.

Reviewed answer: $v\left(t\right)=3t^2-12t+9$ m/s and $a\left(t\right)=6t-12$ m/s²; at 1 s, v=0 and a=−6.

Takeaway: Zero velocity does not imply zero acceleration.

Authored solution to an added practice problem.

1. $(1+\cos x)/\sin x-\cot x=1/\sin x=\csc x$.

   Why: The two cos/sin terms cancel without using a quotient rule.

2. Differentiate the simplified csc x.

   Why: The known special derivative is −csc x cot x.

Reviewed answer: $-\csc x\cot x$, where sin x≠0.

Takeaway: A useful simplification can keep the problem within familiar rules.

Authored solution to an added practice problem.

1. $f(a+h)-f\left(a\right)=2ah+h^2+3h$.

   Why: Expand (a+h)² and cancel identical terms.

2. Divide by nonzero h to get $2a+h+3$, then take h→0.

   Why: The derivative's limit leaves 2a+3.

Reviewed answer: $f\prime \left(a\right)=2a+3$.

Takeaway: The power rule is consistent with the defining limit.

Authored solution to an added practice problem.

1. The table gives $f\prime \left(a\right)=2>0$ and $f\prime \left(b\right)=-1<0$.

   Why: These derivative values describe instantaneous tangent slopes; they do not establish monotonicity throughout a neighborhood.

2. $f\prime \left(c\right)=0$ means a horizontal tangent, but alone does not classify a maximum or minimum.

   Why: A stationary point can be an extremum or neither.

Reviewed answer: Positive tangent slope 2 at a; negative tangent slope −1 at b; horizontal tangent at c.

Takeaway: A zero derivative needs surrounding information before classification.

Authored solution to an added practice problem.

1. For $f\left(x\right)=\sqrt{x}$, $f\left(4\right)=2$ and $f\prime \left(4\right)=1/4$.

   Why: Choose a nearby input with an exact value.

2. The tangent approximation is $L\left(x\right)=2+(x-4)/4$, so $L\left(4.1\right)=2.025$.

   Why: The derivative estimates the small output change from the input change.

Reviewed answer: $\sqrt{4.1}\approx 2.025$.

Takeaway: A tangent line provides a local estimate, not an exact value.

Authored solution to an added practice problem.

1. For $f=6e^x-\sin x$, $f\prime =6e^x-\cos x$ and $f\prime \prime =6e^x+\sin x$.

   Why: The exponential stays fixed while sine and cosine cycle.

2. Differentiate once more to get $6e^x+\cos x$.

   Why: Count the number of derivative passes carefully.

Reviewed answer: $f\prime \prime \prime \left(x\right)=6e^x+\cos x$.

Takeaway: Write intermediate derivatives to track signs and order.

Authored solution to an added practice problem.

1. $x^{1/3}\to 0$ as x→0, equal to f(0).

   Why: The cube root is continuous on the real line.

2. At zero the difference quotient is $h^{1/3}/h=h^{-2/3}$ for nonzero h, which tends to +∞.

   Why: A finite derivative is required for ordinary differentiability.

Reviewed answer: Continuous at 0; no finite derivative there.

Takeaway: An infinite slope is a vertical tangent, not a finite derivative.

Authored solution to an added practice problem.

1. Write f as $x^{-2}$, then differentiate twice.

   Why: The power rule handles reciprocal powers.

2. For positive x, the first derivative is negative and the second positive.

   Why: The graph falls, but its negative slopes become less steep.

Reviewed answer: $f\prime \left(x\right)=-2/x^3<0$ and $f\prime \prime \left(x\right)=6/x^4>0$ for x>0.

Takeaway: A decreasing graph can be concave up.

Authored solution to an added practice problem.

1. For $f=x^3-2x+2$, $f\prime =3x^2-2$. At 0, f=2 and f′=−2, so x₁=1.

   Why: The first tangent intercept moves to 1.

2. At 1, f=1 and f′=1, so x₂=0.

   Why: The next tangent returns to the starting point.

3. The same two steps recur, so these guesses do not approach a root.

   Why: Newton's method is an iteration, not a guarantee of convergence.

Reviewed answer: $x_1=1$, $x_2=0$; these guesses repeat.

Takeaway: Valid denominators do not guarantee convergence.

Authored solution to an added practice problem.

1. The exponent is −3, so the coefficient becomes −3 and the new exponent is −4.

   Why: Reducing an exponent means subtracting one, not changing its sign.

2. Thus $d\left(x^{-3}\right)/dx=-3x^{-4}$, not $3x^{-2}$.

   Why: Both the coefficient and exponent in the proposed derivative are wrong.

Reviewed answer: Correct derivative: $-3x^{-4}$, for x≠0.

Takeaway: The same rule applies to negative exponents.