第12类RD Sharma解决方案-第18章，最大值和最小值–练习18.4

问题1.在给定的时间间隔内找到以下函数的绝对最大值和绝对最小值：

(ⅰ)F(X)= 4× – X在^二分之二[-2，9/2]

解决方案：

Given: f(x) = 4x – x²/2 in [-2, 9/2] 

On differentiation, we get,

f'(x) = 4 – x

For local maxima and local minima we have f'(x) = 0 

4 – x = 0

⇒ x = 4

Let’s evaluate the value of f at the critical point x=4 and at the interval [-2, 9/2]  

f(4) = 4(4) – 4²/2 = 16 – 16/2 = 16 – 8 = 8

f(-2) = 4(-2) – (-2)²/2 = -8 – 4/2 = -8 – 2 = -10

f(9/2) = 4(9/2) – (9/2)²/2 = 18 – 81/8 = 18 – 10.125 = 7.875

Hence, the absolute maximum value of f in [-2, 9/2] is 8 at x = 4 and 

the absolute minimum value of f in [-2, 9/2] is -10 at x = -2.

(ii)f [x] = [x，1] ² +3 in [-3，1]

解决方案：

Given: f(x) = (x – 1)²+3 in [-3, 1]

On differentiation, we get,

f'(x) = 2(x – 1)

For local maxima and local minima we have f'(x) = 0 

2(x – 1) = 0

⇒ x = 1

Let’s evaluate the value of f at the critical point x = 1 and at the interval [-3, 1] 

f(1) = (1 – 1)² + 3 = 3

f(−3) = (-3 – 1)² + 3 = 16 + 3 = 19

Hence, the absolute maximum value of f in [-3, 1] is 19 at x = -3 and 

the absolute minimum value of f in [-3, 1] is 3 at x = 1.

(iii)f(x)= [0，3]中的3x ⁴ – 8x ³ + 12x ² – 48x + 25

解决方案：

Given: f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25 in [0,3]

On differentiation, we get,

f’(x) = 12x³ – 24x² + 24x – 48

= 12(x³ – 2x² + 2x – 4)

= 12(x – 2)(x² + 2)

Now, for local minima and local maxima we have f′(x) = 0

x = 2 or  x² + 2 = 0 having no real roots

therefore we consider only x = 2 ∈ [0, 3]

Let’s evaluate the value of f at the critical point x = 2 and at the interval [0, 3]

f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48(2) + 25 

= 3(16) – 8(8) + 12(4) – 96 + 25

= 48 – 64 + 48 – 96 + 25 

= -39

f(0) = 3(0)⁴ – 8(0)³ + 12(0)² – 48(0) + 25 = 25

f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48(3) + 25 

= 3(81) – 8(27) + 12(9) – 144 + 25

= 243 – 216 + 108 – 144 + 25

= 16

Hence, the absolute maximum value of f in [0, 3] is 25 at x = 0 and 

the absolute minimum value of f in [0, 3] is -39 at x = 2. 

(iv) $f(x)=(x-2) \sqrt{x-1}$在[1，9]中

解决方案：

Given: $f(x)=(x-2) \sqrt{x-1}$in [1, 9]

On differentiation, we get,

$f'(x) = \sqrt{x-1} + \frac {x-2} {2\sqrt {x-1}}$

Now, for local minima and local maxima we have f′(x) = 0

$\sqrt{x-1} + \frac {x-2} {2\sqrt {x-1}}=0$

= $\frac {2(x-1) + {x-2}} {2\sqrt {x-1}}=0$

= $\frac {3x-4} {2\sqrt {x-1}}=0$

= x = 4/3

Let’s evaluate the value of f at the critical point x = 4/3 and at the interval [1, 9] 

$f(1)=(1-2)\sqrt {1-1}=(-1)\sqrt {0}=0$

$f(\frac 4 3)= (\frac 4 3 -2)\sqrt {\frac 4 3 -1}=\frac {4-6} {3\sqrt3}=\frac {-2} {3\sqrt3}=\frac {-2\sqrt3} 9$

$f(9)=(9-2)\sqrt{9-1}=7\sqrt8=14\sqrt2$

Hence, the absolute maximum value of f in [1, 9] is 14√2 at x = 9 and 

the absolute minimum value of f in [1, 9] is $\frac {-2\sqrt3} 9$ at x = 4/3

问题2.在区间[1，3]中找到2x ³ – 24x + 107的最大值。在[-3，-1]中找到相同函数的最大值。

解决方案：

Let f(x) = 2x³ – 24x + 107 

∴ f'(x) = 6x² – 24 = 6(x² – 4) 

Now, for local maxima and local minima we have f'(x) = 0

⇒ 6(x² – 4) = 0

⇒ x² = 4

⇒ x = ±2

We first consider the interval [1, 3].

Let’s evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at  the interval [1, 3].

f(2) = 2(2)³ – 24(2) + 107 = 75

f(1) = 2(1)³ – 24(1) + 107 = 85

f(3) = 2(3)³ – 24 (3) + 107 = 89

Hence, the absolute maximum value of f in the interval [1, 3] is 89 occurring at x = 3,

Next, we consider the interval [– 3, – 1].

Evaluate the value of f at the critical point x = – 2 ∈ [-3, -1]

f(−3) = 2(-3)³ – 24(-3) + 107 = 125

f(-2) = 2(-2)³ – 24(-3) + 107 = 139

f(-1) = 2(-1)³ – 24(-2) + 107 = 129

Hence, the absolute maximum value of f is 139 when x = -2.

问题3.求出由f(x)= cos ² x + sinx，x∈[0， π ]给出的函数f的绝对最大值和最小值。

解决方案：

Given: f(x) = cos²x + sinx, x ∈ [0, π]

On differentiation, we get,

f′(x) = 2cosx(−sinx) + cosx

= −2sinxcosx + cosx

Now, for local minima and local maxima we have f′(x) = 0 

−2sinxcosx + cosx = 0

⇒ cosx(−2sinx + 1) = 0

⇒ sinx = 1​/2 or cos x = 0

⇒ x = π/6​, π/2​ as x ∈ [0, π]

Let’s evaluate the value of f at the critical points x=6/π and x = 2/π and at the interval [0, π]

f(π/6​) = cos²π/6 + sinπ/6 = (√3/2​​)²+1/2 ​= 5/4​

f(π/2​) = cos²π/2​ + sinπ/2​ = 0 + 1 = 1

f(0) = cos²0 + sin0 = 1 + 0 = 1

f(π) = cos²π + sinπ = (−1)² + 0 = 1

Hence, the absolute maximum value of f in [0, π] is 5/4 at x = π/6​   and

the absolute minimum value of f in [0, π] is 1 at x = 0, π/2​, π.

问题4.求出函数f的绝对最大值和最小值，该函数由f(x)= 12x ^4/3 ?? – 6x ^1/3 ，x∈[−1，1]给出。

解决方案：

Given: f(x) = 12x^4/3​ – 6x^1/3​, x ∈ [−1, 1].

On differentiation, we get,

$f'(x)=16x^{\frac 1 3}-\frac 2 {x^{\frac 2 3}}=\frac {2(8x-1)} {x^{\frac 2 3}}$

Now, for local minima and local maxima we have f′(x) = 0

$\frac {2(8x-1)} {x^{\frac 2 3}}=0$

⇒ x = 1/8

Let’s evaluate the value of f at the critical points x = 1/8 and at the interval [-1, 1]

f(1/8) = 12(1/8)^4/3 – 6(1/8)^1/3 = -9/4

f(−1) = 12(−1)^4/3– 6(−1)^1/3 ​= 18

f(1) = 12(1)^4/3– 6(1)^1/3 ​= 6

Hence, the absolute maximum value of f in [-1, 1] is 18 at x = -1 and 

the absolute minimum value of f in [-1, 1] is -9/4 at x = 1/8.

问题5.在区间[1，5]中找到f(x)= 2x ³ – 15x ² + 36x + 1给定的函数f的绝对最大值和最小值。

解决方案：

Given: f(x) = 2x³ – 15x²+ 36x + 1 in the interval [1, 5]

On differentiation, we get,

f′(x) = 6x² – 30x + 36 = 6(x² – 5x + 6) = 6(x – 2)(x – 3)

Now, for local minima and local maxima we have f′(x) = 0

6(x – 2)(x – 3) = 0

⇒ x = 2 and x = 3 

Let’s evaluate the value of f at the critical points x = 2 and x = 3 and at the interval [1, 5]

f(1) = 2(1)³ – 15(1)² + 36(1) + 1 = 24

f(2) = 2(2)³ – 15(2)² + 36(2) + 1 = 29

f(3) = 2(3)³ – 15(3)² + 36(3) + 1 = 28

f(5) = 2(5)³ – 15(5)² + 36(5) + 1 = 56

Hence, the absolute maximum value of f in [1, 5] is 56 at x = 5 and 

the absolute minimum value of f in [1, 5] is 24 at x = 1.