3 is a local maximum value and f (±1

4− 3

11. (a) f (x) = x⁴−2x²+3⇒ f⁰(x) = 4x³−4x = 4x(x²−1) = 4x(x+1)(x−1).

(b) f changes from increasing to decreasing at x = 0 and from decreasing to increasing at x = −1 and x = 1. Thus, f(0) = 3 is a local maximum value and f (±1) = 2 are local minimum values.

(c) f⁰⁰(x) = 12x²− 4 = 12(x²−¹₃) = 12((x +√¹

3)(x−^√1₃)).

f⁰⁰(x) > 0⇔ x < −^√1₃ or x >√¹

3 and f⁰⁰(x) < 0⇔ −^√1₃< x < √¹ 3. Thus, f is concave upward on (−∞, −^√₃³) and (

√3

3 ,∞) and concave down- ward on (−^√₃³,^√₃³). There are inflection points at (±^√₃³,²²₉).

12. (a) f (x) = _x2^x+3² ⇒ f⁰(x) = ^(x2+3)(2x)_(x2+3)^−x22(2x) = _(x2^6x+3)². The denominator is positive so the sign of f⁰(x) is determined by the sign of x. Thus, f⁰(x) > 0⇔ x > 0 and f⁰(x) < 0⇔ x < 0. So f is increasing on (0, ∞) and f is decreasing on (−∞, 0).

(b) f changes from decreasing to increasing at x = 0. Thus, f (0) = 0 is a local minimum value.

(c) f⁰⁰(x) = ^(x2+3)2(6)_[(x^−6x·2(x2+3)²]^22+3)(2x) =^6(x2+3)[x_(x2+3)^2+34−4x2]

= ⁶⁽³_(x2^−3x+3)²³⁾ =−18(x+1)(x−1) (x²+3)³ .

f⁰⁰(x) > 0 ⇔ −1 < x < 1 and f⁰⁰(x) < 0 ⇔ x < −1 or x > 1. Thus, f is concave upward on (−1, 1) and concave downward on (−∞, −1) and (1,∞). There are inflection points at (±1,¹₄).

16. (a) f (x) = x²ln x⇒ f⁰(x) = x²(1/x) + (ln x)(2x) = x + 2x ln x = x(1 + 2 ln x). The domain of f is (0,∞), so the sign of f⁰ is determined solely by the factor 1 + 2 ln x. f⁰(x) > 0 ⇔ ln x > −¹₂ ⇔ x > e⁻¹² [≈ 0.61]

and f⁰(x) < 0⇔ 0 < x < e⁻¹² . So f is increasing on (e⁻¹² ,∞) and f is decreasing on (0, e⁻¹² ).

1

(b) f changes from decreasing to increasing at x = e⁻¹² . Thus, f (e⁻¹² ) = (e⁻¹² )²ln(e⁻¹² ) = e⁻¹(⁻¹₂ ) =−_2e¹ [≈ −0.18] is a local minimum value.

(c) f⁰(x) = x(1+2 ln x)⇒ f⁰⁰(x) = x(2/x)+(1+2 ln x)·1 = 2+1+2 ln x = 3+2 ln x. f⁰⁰(x) > 0⇔ 3+2 ln x > 0 ⇔ ln x > −3/2 ⇔ x > e^−3/2[≈ 0.22].

Thus, f is concave upward on (e^−3/2,∞) and f is concave downward on (0, e^−3/2). f (e^−3/2) = (e^−3/2)²ln e^−3/2 = e⁻³(−3/2) = _2e⁻³3 [≈ −0.07].

There is a point of inflection at (e^−3/2, f (e^−3/2)) = (e^−3/2,_2e⁻³₃).

20. f (x) = _x2^x+4 ⇒ f⁰(x) = ^(x2+4)_(x2^·1−x(2x)+4)² = _(x⁴2^−x+4)²² =^(2+x)(2_(x2+4)^−x)2 .

First Derivative Test: f⁰(x) > 0⇒ −2 < x < 2 and f⁰(x) < 0⇒ x > 2 or x <−2. Since f⁰ changes from positive to negative at x = 2, f (2) = ¹₄ is a local maximum value; and since f⁰ changes from negative to positive at x =−2, f(−2) = −¹₄ is a local minimum value.

Second Derivative Test: f⁰⁰(x) =^{(x2+4)2(−2x)−(4−x}_[(x2+4)²²]^{)2·2(x2+4)(2x)}

= ^{−2x(x2+4)[(x}_(x2+4)^{2+4)+2(44 −x2)]} =^−2x(12−x_(x2+4)³²⁾. f⁰(x) = 0⇔ x = ±2.

f⁰⁰(−2) = ₁₆¹ > 0⇒ f(−2) = −¹₄ is a local minimum value.

f⁰⁰(2) =−₁₆¹ < 0⇒ f(2) = ¹₄ is a local maximum value.

Preference: Since calclating the second derivative is fairly difficult, the First Derivative Test is easier to use for this function.

21. f (x) = x +√

1− x ⇒ f⁰(x) = 1 +¹₂(1− x)^−1/2(−1) = 1 −₂^√1_1−x. Note that f is defined for 1− x ≥ 0; that is, for x ≤ 1. f⁰(x) = 0⇒ 2√

1− x = 1⇒√

1− x = ¹₂ ⇒ 1 − x = ¹₄ ⇒ x = ³₄. f⁰ does not exist at x = 1, but we can’t have a local maximum of minimum at an endpoint.

First Derivative Test: f⁰(x) > 0⇒ x < ³₄ and f⁰(x) < 0⇒ ³₄ < x < 1.

Since f⁰ changes from positive to negative at x = ³₄, f (³₄) = ⁵₄ is a local maximum value.

Second Derivative Test: f⁰⁰(x) =−¹₂(−¹₂)(1− x)^−3/2(−1) = −₄₍^√₁¹_−x)3. f⁰⁰(³₄) =−2 < 0 ⇒ f(³₄) =⁵₄ is a local maximum value.

Preference: The First Derivative Test may be slightly easier to apply in this case.

22. (a) f (x) = x⁴(x− 1)³⇒ f⁰(x) = x⁴· 3(x − 1)²+ (x− 1)³· 4x³

= x³(x− 1)²[3x + 4(x− 1)] = x³(x− 1)²(7x− 4).

The critical numbers are 0, 1, and ⁴₇.

(b) f⁰⁰(x) = 3x²(x− 1)²(7x− 4) + x³· 2(x − 1)(7x − 4) + x³(x− 1)²· 7

= x²(x− 1)[3(x − 1)(7x − 4) + 2x(7x − 4) + 7x(x − 1)].

Now f⁰⁰(0) = f⁰⁰(1) = 0, so the Second Derivative Test gives no informa- tion for x = 0 or x = 1.

2

f⁰⁰(⁴₇) = (⁴₇)²(⁴₇− 1)[0 + 0 + 7(⁴₇)(⁴₇− 1)] = (⁴₇)²(−³₇)(4)(−³₇) > 0, so there is a local minimum at x = ⁴₇.

(c) f⁰ is positive on (−∞, 0), negative on (0,⁴₇), positive on (⁴₇, 1), and positive on (1,∞). So f has a local maximum at x = 0, a local minimum at x =⁴₇, and no local maximum or minimum at x = 1.

41. (a) C(x) = x^1/3(x + 4) = x^4/3+ 4x^1/3

⇒ C⁰(x) = ⁴₃x^1/3+⁴₃x^−2/3 =⁴₃x^−2/3(x + 1) =^4(x+1)

3√³ x² . C⁰(x) > 0 if−1 < x < 0 or x > 0 and C⁰(x) < 0 for x <−1, so C is increasing on (−1, ∞) and C is decreasing on (−∞, −1).

(b) C(−1) = −3 is a local minimum value.

(c) C⁰⁰(x) = ⁴₉x^−2/3−⁸₉x^−5/3 =⁴₉x^−5/3(x− 2) = ^4(x₉√3⁻²⁾

x⁵ .

C⁰⁰(x) < 0 for 0 < x < 2 and C⁰⁰(x) > 0 for x < 0 and x > 2, so C is concave downward on (0, 2) and concave upward on (−∞, 0) and (2, ∞).

There are inflection points at (0, 0) and (2, 6√³

2)≈ (2, 7.56).

(d)

76. (a) Let f (x) = e^x− 1 − x ⇒ f⁰(x) = e^x− 1 for x ≥ 0.

Because f⁰(x)≥ 0 for x ≥ 0 ⇒ f is increasing on (0, ∞)

⇒ f(x) ≥ f(0) = 0 for x ≥ 0.

⇒ e^x≥ 1 + x for x ≥ 0.

(b) Let g(x) = e^x− 1 − x − ¹₂x²⇒ g⁰(x) = e^x− 1 − x for x ≥ 0.

By (a), g⁰(x)≥ 0 for x ≥ 0 ⇒ g is increasing on (0, ∞)

⇒ g(x) ≥ g(0) = 0 for x ≥ 0.

⇒ e^x≥ 1 + x +¹₂x² for x≥ 0.

(c) As n = 1, by (a), we know e^x≥ 1 + x for x ≥ 0.

As n = k, e^x≥ 1 + x +¹₂x²+ ... + _k!¹x^k is holded for x≥ 0.

As n = k + 1, Let h(x) = e^x− (1 + x +¹₂x²+ ... +_(k+1)!¹ x^(k+1))

3

h⁰(x) = e^x− (1 + x +¹₂x²+ ... + _k!¹x^k)≥ 0 for x ≥ 0.

i.e. h is increasing for x≥ 0, then h(x)≥ h(0) = 0 for x ≥ 0

⇒ e^x≥ 1 + x +¹₂x²+ ... + _(k+1)!¹ x^(k+1) for x≥ 0.

By induction, we know e^x≥ 1 + x +¹₂x²+ ... +_n!¹xⁿ for x≥ 0 any positive integer n.

4