The area of the rectangle under the curve from −x to x is A(x

[Ch4 Problems Plus]

1. Let y = f (x) = e^−x2. First Note that f (x) is an even function, hence symmetric with respect to the y-axis. Therefore the two vertices of the rect- angle we are considering are f (x) and f (−x) for some x. The area of the rectangle under the curve from −x to x is A(x) = 2xe^−x2 where x ≥ 0.

We maximize A(x): A⁰(x) = 2e^−x2−4x²e^−x2 = 2e^−x2(1−2x²) = 0 ⇒ x = ^√1

2. This gives a maximum since A⁰(x) > 0 for 0 ≤ x < ^√1

2 and A⁰(x) < 0 for x > ^√1₂.

We next determine the points of inflection of f (x). Notice that f⁰(x) =

−2xe^−x2 = −A(x). So f⁰⁰(x) = −A⁰(x) and hence, f⁰⁰(x) < 0 for −^√1

2 <

x < ^√1

2 and f⁰⁰(x) > 0 for x < −^√1

2 and x > ^√1

2. So f (x) changes concavity at x = ±^√1

2, and the two vertices of the rectangle of largest area are at the inflection points.

2. Let f (x) = sin x − cos x on [0, 2π] since f has period 2π.

f⁰(x) = cos x + sin x = 0 ⇔ cos x = − sin x ⇔ tan x = −1 ⇔ x = ^3π₄ or

7π

4 . Evaluation f at its critical numbers and endpoints, we get f (0) = −1, f (^3π₄ ) = √

2, f (^7π₄ ) = −√

2, and f (2π) = −1. So f has absolute maximum value √

2 and absolute minimum value −√

2. Thus −√

2 ≤ sin x − cos x ≤

√2 ⇒ | sin x − cos x| ≤ √ 2.

5. y = ^{sin x}_x ⇒ y⁰ = x cos x−sin x

x² ⇒ y⁰⁰ = ^−x2sin x−2x cos x+2 sin x

x³ .

If (x, y) is an inflection point, then y⁰⁰ = 0

⇒ (2 − x²) sin x = 2x cos x ⇒ (2 − x²)²sin²x = 4x²cos²x

⇒ (2 − x²)²sin²x = 4x²(1 − sin²x) ⇒ (4 − 4x²+ x⁴) sin²x = 4x²− 4x²sin²x

⇒ (4 + x⁴) sin²x = 4x² ⇒ (x⁴+ 4)^sin_x²2^x = 4

⇒ y²(x⁴+ 4) = 4 since y = ^{sin x}_x .

6. y = 1−x² ⇒ y⁰ = −2x. Let P (a, 1−a²) be the point of contact. The slope of the tangent line at P is −2a, hence the equation of the tangent line at P is y − (1 − a²) = (−2a)(x − a) ⇒ y − 1 + a² = −2a + 2a² ⇒ y = −2ax + a²+ 1.

To find the x-intercept, put y = 0: 2ax = a²+ 1 ⇒ x = ^a2_2a⁺¹. To find the y-intercept, put x = 0: y = a²+ 1.

Therefore, the area of the triangle is ¹₂(^a2_2a⁺¹)(a²+ 1) = ^(a2_4a⁺¹⁾². We minimize the function A(a) = ^(a2_4a⁺¹⁾², a > 0.

A⁰(a) = ^{(4a)2(a2+1)(2a)−(a}_16a2 ^2+1)2(4) = ^(a2+1)[4a_4a²2^−(a2+1)] = ^(a2+1)(3a_4a2 ²⁻¹⁾.

A⁰(a) = 0 when 3a² − 1 = 0 ⇒ a = ^√1₃. A⁰(a) < 0 for a < ^√1₃ and A⁰(a) > 0 for a > ^√1

3. So by the First Derivative Test, there is an absolute minimum when a = ^√1

3. The required point is (^√1

3,²₃) and the corresponding minimum 1

area is A(^√1

3) = ⁴

√ 3 9 .

7. Let L = lim_x→0 ^ax2+sin bx+sin cx+sin dx

3x²+5x⁴+7x⁶ . Now L has the indeterminate form of type ⁰₀, so we can apply l’Hospital’s Rule.

L = limx→0 2ax+b cos bx+c cos cx+d cos dx

6x+20x³+42x⁵ . The denominator approaches 0 as x → 0, so the numerator must also approach 0 because the limit exist. Now the nu- merator approaches 0 + b + c + d, so b + c + d = 0.

Apply l’Hospital’s Rule again. L = limx→0 2a−b²sin bx−c²sin cx−d²sin dx

6+60x²+210x⁴ = ^2a−0₆₊₀ =

2a

6 , which must equal 8. ^2a₆ = 8 ⇒ a = 24.

Thus, a + b + c + d = a + (b + c + d) = 24 + 0 = 24.

9. Differentiating x² + xy + y² = 12 implicity with respect to x gives 2x + y + x_dx^dy + 2y^dy_dx = 0. So ^dy_dx = −^2x+y_x+2y.

At a highest of lowest point, ^dy_dx = 0 ⇔ y = −2x. Substituting −2x for y in the original equation gives x²+ x(−2x) + (−2x)² = 12, so 3x² = 12 and x = ±2.

If x = 2, then y = −2x = −4, and if x = −2, then y = 4. Thus, the highest and lowest points are (−2, 4) and (2, −4).

10. y = cx³ + e^x ⇒ y⁰ = 3cx² + e^x ⇒ y⁰⁰ = 6cx + e^x. The curve will have inflection points when y⁰⁰ changes sign. y⁰⁰ = 0 ⇒ −6cx = e^x, so y⁰⁰ will change sign when the line y = −6cx intersects the curve y = e^x (but is not tangent to it).

Note that if c = 0, the curve is just y = e^x, which has no inflection point.

For c > 0, y = −6cx will intersect y = e^x once, so y = cx³+ e^x will have one inflection point.

For c < 0, y = −6cx can intersect the curve y = e^x in two points (two in- flection points), be tangent to it (no inflection point), or not intersect it (no inflection point). Suppose y = −6cx is the tangent line at (a, e^a). The tan- gent line has slope e^a, and it is y−e^a= e^a(x−a) ⇒ y = e^ax−ae^a+e^a= −6cx.

So −ae^a+ e^a = 0 ⇒ a = 1. Thus, the slope is e. Next, suppose y = −6cx intersect y = e^x in two points, the line y = −6cx must have slope greater than e, so −6c > e ⇒ c < −^e₆.

Therefore, the curve y = cx³+ e^x will have one inflection point if c > 0 and two inflection points if c < −₆^e.

16. If L = lim_x→∞(^x+a_x−a)^x, then L has the indeterminate form 1^∞. So ln L = lim_x→∞ln(^x+a_x−a)^x = lim_x→∞x ln(^x+a_x−a) = lim_x→∞ln(x+a)−ln(x−a)

1/x

= lim_x→∞

1 x+a−_x−a¹

−1/x² by l’Hospital’s Rule

= lim_x→∞[(x−a)−(x+a)

(x+a)(x−a) · ^−x₁²] = lim_x→∞x^2ax2−a²² = lim_x→∞1−a^2a2/x² = 2a 2

Hence, ln L = 2a, so L = e^2a. From the original equation, we want L = e¹ ⇒ 2a = 1 ⇒ a = ¹₂.

18. Let the circle have radius r, so |OP | = |OQ| = r, where O is the center of the circle. Now ∠P OR has measure ¹₂θ, and ∠OP R is a right angle, so tan¹₂θ = ^{|P R|}_r and the area of 4OP R is ¹₂|OP ||P R| = ¹₂r²tan¹₂θ. The area of the sector cut by OP and OR is ¹₂r²(¹₂θ) = ¹₄r²θ. Let S be the intersection of P Q and OR. Then sin¹₂θ = ^{|P S|}_r and cos ¹₂θ = ^|OS|_r , and the area of 4OSP is ¹₂|OS||P S| = ¹₂(r cos¹₂θ)(r sin¹₂θ) = ¹₂r²sin¹₂θ cos¹₂θ = ¹₄r²sin θ.

So B(θ) = 2(¹₂r²tan¹₂θ − ¹₄r²θ) = r²(tan¹₂θ −¹₂θ), and A(θ) = 2(¹₄r²θ − ¹₄r²sin θ) = ¹₂r²(θ − sin θ), Thus, lim_θ→0^{+ A(θ)}_B(θ) = lim_θ→0⁺

1

2r²(θ−sin θ)

r²(tan¹₂θ−¹₂θ) = lim_θ→0⁺ = ^{θ−sin θ}

2(tan¹₂θ−¹₂θ)

= lim_θ→0⁺ ₂₍1^{1−cos θ}

2sec^{2 1}₂θ−¹₂) by l’Hospital’s Rule

= lim_θ→0⁺ _sec^{1−cos θ}_{2 1}

2θ−1 = lim_θ→0⁺ _tan^{1−cos θ}_{2 1}

2θ

= lim_θ→0⁺ _2(tan1^{sin θ}

2θ)(sec^{2 1}₂)¹₂ by l’Hospital’s Rule

= lim_θ→0^{+ sin θ cos}_sin1^{3 12θ}

2θ = lim_θ→0^{+ (2 sin}

1

2θ cos¹₂θ) cos^{3 1}₂θ

sin¹₂θ = 2 lim_θ→0⁺cos⁴(¹₂θ) = 2(1)⁴ = 2.

Remark.(1)The area of a sector of circle: A = ¹₂r²θ (2) sec²θ − 1 = tan²θ

(3) sec θ = _{cos θ}¹

(4) sin 2θ = 2 sin θ cos θ

20. A straight line intersects the curve y = f (x) = x⁴+ cx³+ 12x²− 5x + 2 in four distant points if and only if the graph of f has two inflection points.

f⁰(x) = 4x³+ 3cx²+ 24x − 5 and f⁰⁰(x) = 12x²+ 6cx + 24.

f⁰⁰(x) = 0 ⇔ x = ^−6c±

√

(6c)²−4(12)(24)

2(12) . There are two distinct roots for f⁰⁰(x) = 0 (and hence two inflection points) if and only if the discriminant is positive; that is, 36c²− 1152 > 0 ⇔ |c| >√

32.

Thus, the desired values of c are c < −4√

2 or c > 4√ 2.

3