第12类RD Sharma解决方案–第18章，最大值和最小值–练习18.2

使用一阶导数检验，找到以下函数的局部最大值或局部最小值的点。此外，视情况而定，找到局部最大值或局部最小值：

问题1. f(x)=(x – 5) ⁴

解决方案：

Given function 

f(x) = (x – 5)⁴

Now, differentiate the given function w.r.t. x

f ‘(x) = 4(x-5)³

Now, for local maxima and minima  

Put f ‘(x) = 0

⇒ 4(x – 5)³ = 0

⇒ x – 5 = 0

⇒ x = 5

So, at x = 5, f'(x) changes from negative to positive. Hence, x = 5 is the point of local minima 

So, the minimum value is f(5) = (5 – 5)⁴ = 0

问题2。f(x)= x ³ – 3x

解决方案：

Given function 

f(x) = x³– 3x

Now, differentiate the given function w.r.t. x

f ‘(x) =  3x²– 3

Now, for local maxima and minima  

Put f ‘(x) = 0

⇒ 3x²– 3 = 0

⇒ x = ±1

Now, again differentiating f'(x) function w.r.t. x

f “(x) = 6x

Put x = 1 in f”(x)

f “(1)= 6 > 0

So, x = 1 is point of local minima  

Put x = -1 in f”(x)

f “(-1)= -6 < 0

So, x = -1 is point of local maxima  

So, the minimum value is f(1) = x³– 3x = 1³ – 3 = -2

and the maximum value is f(-1) = x³– 3x = (-1)³ – 3(-1) = 2                                  

问题3。f(x)= x ³ (x – 1) ²

解决方案：

Given function 

f(x) = x³(x – 1)²

Now, differentiate the given function w.r.t. x

f ‘(x) = 3x² (x- 1)² + 2x³(x- 1)

= (x – 1) (3x²(x – 1) + 2x³)  

= (x – 1) (3x³ – 3x² + 2x³)

= (x – 1) (5x³ – 3x²)

= x²(x – 1) (5x – 3)

Now, for all maxima and minima,

Put f ‘(x) = 0

= x²(x – 1) (5x- 3) = 0

x = 0, 1, 3/5

So, at x = 3/5, f'(x) changes from negative to positive. Hence, x = 3/5 is a point of minima

So, the minimum value is f(3/5) = (3/5)³(3/5 – 1)² = 108/3125

At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of maxima.

So, the maximum value is f(1) = (1)³(1 – 1)² = 0

问题4. f(x)=(x – 1)(x + 2) ²

解决方案：

Given function 

f(x) = (x – 1)(x + 2)²

Now, differentiate the given function w.r.t. x

f ‘(x) = (x + 2)² + 2(x – 1)(x + 2)

= (x+ 2) (x+ 2 + 2x – 2)

 =(x + 2) (3x)

Now, for all maxima and minima,

Put f ‘(x) = 0

⇒ (x + 2) (3x) = 0

x = 0,-2

So, at x = -2, f(x) changes from positive to negative. Hence, x = -2 is a point of Maxima

So, the maximum value is f(-2) = (-2 – 1)(-2 + 2)² = 0

At x = 0, f ‘(x) changes from negative to positive. Hence, x = 0 is point of Minima.

So, the minimum value is f(0) = (0 – 1)(0 + 2)² = -4

问题5. f(x)=(x – 1) ³ (x + 1) ²

解决方案：

Given function 

f(x) = (x – 1)³ (x + 1)²

Now, differentiate the given function w.r.t. x

f ‘(x) = 3(x – 1)² (x + 1)² + 2(x – 1)³ (x + 1)

= (x – 1)² (x + 1) {3(x + 1) + 2(x – 1)}

= (x – 1)² (x + 1) (5x + 1)

Now, for local maxima and minima,

Put f ‘(x) = 0

⇒ (x – 1)² (x + 1) (5x + 1) = 0

⇒ x = 1, -1, -1/5

So, at x = -1, f ‘(x) changes from positive to negative. Hence, x = -1 is point of maxima

So, the maximum value is f(-1) = (-1 – 1)³ (-1 + 1)² = 0

At x = -1/5, f ‘(x) changes from negative to positive so x= -1/5 is point of minima

So, the minimum value is f(-1/5) = (-1/5 – 1)³ (-1/5 + 1)² = -3456/3125

问题6. f(x)= x ³ – 6x ² + 9x +15

解决方案：

Given function 

f(x) = x³ – 6x² + 9x + 15

Now, differentiate the given function w.r.t. x

f ‘(x) = 3x² – 12x + 9

= 3 (x² – 4x + 3)

= 3 (x – 3) (x – 1)

Now, for local maxima and minima,

 Put f ‘(x) = 0

⇒ 3 (x – 3) (x – 1) – 0

⇒ x = 3, 1

At x = 1, f'(x) changes from positive to negative. Hence, x = 1 is point of local maxima

So, the maximum value is f(1) = (1)³ – 6(1)² + 9(1) + 15 = 19

At x = 3, f'(x) changes from negative to positive. Hence, x = 3 is point of local minima

So, the minimum value is f(x) = (3)³ – 6(3)² + 9(3) + 15 = 15

问题7. f(x)= sin2x，0

解决方案：

Given function 

f(x) = sin2x, 0 < x, π

Now, differentiate the given function w.r.t. x

f'(x) = 2 cos 2x

Now, for local maxima and minima,

Put f'(x) = 0

⇒ 2x = $\frac{π}{2},\frac{3π}{2}$

⇒ x = $\frac{π}{4},\frac{3π}{4}$

At x = π​/4, f'(x) changes from positive to negative. Hence, x = π​/4, Is point of local maxima

So, the maximum value is f(π​/4) = sin2(π​/4) = 1

At x = 3π​/4, f'(x) changes from negative to positive. Hence, x = 3π​/4 is point of local minima,

So, the minimum value is f(3π​/4) = sin2(3π​/4) = -1

问题8. f(x)= sin x – cos x，0

解决方案：

Given function 

f(x) = sin x – cos x, 0 < x < 2π

Now, differentiate the given function w.r.t. x

f'(x)= cos x + sin x

Now, for local maxima and minima,

Put f'(x) =0 

cos x = -sin x 

tan x = -1

x = $\frac{3π}{4},\frac{7π}{4}$∈ (0, 2π)

Now again differentiate the given function w.r.t. x

f”(x) = -sin x + cos x

$f"(\frac{3π}{4})=-sin\frac{3π}{4}+cos\frac{3π}{4}=-\frac{1}{√2}-\frac{1}{√2}=-√2$<0

$f"(\frac{7π}{4})=-sin\frac{7π}{4}+cos\frac{7π}{4}=\frac{1}{√2}+\frac{1}{√2}=√2$>0

Therefore, by second derivative test, $x=\frac{3π}{4}$is a point of local maxima 

Hence, the maximum value is $f(\frac{3π}{4})=sin\frac{3π}{4}-cos\frac{3π}{4}=\frac{1}{√2}+\frac{1}{√2}=√2.$

However, $x=\frac{7π}{4}$is a point of local minima 

Hence, the minimum value is $f(\frac{7π}{4})=sin\frac{7π}{4}-cos\frac{7π}{4}=-\frac{1}{√2}-\frac{1}{√2}=-√2$

问题9. f(x)= cos x，0

解决方案：

Given function 

f(x) = cos x, 0< x < π

Now, differentiate the given function w.r.t. x

f'(x) = – sin x

Now, for local maxima and minima,

 Put f ‘(x) – 0

⇒ – sin x = 0

⇒ x = 0, and π

But, these two points lies outside the interval (0, π)

So, no local maxima and minima will exist in the interval (0, π).

问题10。f(x)= sin 2x – x，-π/ 2≤x≤π/ 2

解决方案：

Given function 

f(x) = sin2x – x

Now, differentiate the given function w.r.t. x

f ‘(x) = 2 cos 2x – 1

Now, for local maxima and minima,

Put f'(x) = 0

⇒ 2cos 2x – 1 = 0

⇒ cos 2x = 1/2 = cos π/3

⇒ 2x = π/3, -π/3

⇒ x = $\frac{π}{6},-\frac{π}{6}$

At x = -π/6, f'(x) changes from negative to positive. Hence, x = π/6 is point of local minima.

So, the minimum value is $f(-\frac{π}{6})=\frac{-√3}{2}+\frac{π}{6}$

At x = π/6, f'(x) changes from positive to negative. Hence, x = π/6 is point of local maxima

The maximum value is $f(\frac{π}{6})=\frac{√3}{2}-\frac{π}{6}$

问题11. f(x)= 2sin x – x，-π/ 2≤x≤π/ 2

解决方案：

Given function 

f(x) = 2sin x – x, -π/2≤ x ≤ π/2

Now, differentiate the given function w.r.t. x

f ‘(x) = 2cos x – 1 = 0

Now, for local maxima and minima,

Put f'(x) = 0

⇒ cos x = 1/2 = cos π/3

⇒ x = -π/3, π/3

So, at x = -π/3, f'(x) changes from negative to positive. Hence, x = -π/3 is point of local minima 

So, the minimum value is f(-π/3) = 2sin(-π/3) – (-π/3) = -√3 – π/3

At x = π/3, f'(x) changes from positive to negative. Hence, x = π/3 is point of local minima 

The maximum value is f(π/3) = 2sin(π/3) – (π/3) = √3 – π/3

问题12. f(x)= x $\sqrt{1- x}$ ，x> 0

解决方案：

Given function 

f(x) = x$\sqrt{1- x}$, x > 0

Now, differentiate the given function w.r.t. x

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

= $\frac{2(1-x)-x}{2\sqrt{1- x}}=\frac{2-3x}{2\sqrt{1- x}}$

Now, for local maxima and minima,

Put f'(x) = 0

⇒ $\frac{2-3x}{2\sqrt{1-x}}=0$

⇒ 2 – 3x = 0

⇒ x = 2/3

f “(x) = $\frac{1}{2}[\frac{\sqrt{1-x}(-3)-(2-3x)(\frac{-1}{2\sqrt{1-x}})}{1-x}]$

=$\frac{\sqrt{1-x}(-3)+(2-3x)(\frac{1}{2\sqrt{1-x}})}{2(1-x)}$

= $\frac{-6(1-x)+(2-3x)}{4(1-x)^{3/2}}$

= (3x – 4)/4(1 – x)²

f “(2/3) = $\frac{3(\frac{2}{3})-4}{4(1-\frac{2}{3})^{3/2}}$

= $\frac{2-4}{4(\frac{1}{3})^{3/2}}$

= $\frac{-1}{2(\frac{1}{3})^{3/2}}<0$

Therefore, x = 2/3 is a point of local maxima and the local maximum value of f at x = 2/3 is 

f(2/3) = 2/3(√1/3) = (2√3)/9

问题13. f(x)= x ³ (2x – 1) ³

解决方案：

Given function 

f(x) = x³(2x – 1)³

Now, differentiate the given function w.r.t. x

f'(x) = 3x²(2x – 1)² + 6x³(2x – 1)²

= 3x²(2x – 1)²(2x – 1 + 2x)

= 3x²(4x – 1) 

Now, for local maxima and minima,

Put f'(x) = 0

⇒ 3x²(4x – 1) = 0

⇒ x = 0, 1/4

At x = 1/4, f'(x) changes from negative to positive. Hence, x = 1/4 is the point of local minima,

So, the minimum value is f(1/4)= (1/4)³(2(1/4) – 1)³= -1/512

问题14. f(x)= x / 2 + 2 / x，x> 0

解决方案：

Given function 

f(x) = x/2 + 2/x, x > 0

Now, differentiate the given function w.r.t. x

f'(x) = 1/2 – 2/x², x > 0

Now, for local maxima and minima,

Put f'(x) = 0

⇒ 1/2 – 2/x² = 0

⇒ x² – 4 = 0

⇒ x = 2, -2

At x = 2, f'(x) changes from negative to positive. Hence, x = 2 is point of local minima 

So, the local minimum value is f(2) = 2/2 + 2/2 = 2

问题15. f(x)= 1 /(x ² + 2)

解决方案：

Given function 

f(x) = 1/(x² + 2)

Now, differentiate the given function w.r.t. x

f'(x) = -(2x)/(x² + 2)²

Now, for local maxima and minima,

Put f'(x) = 0  f'(x) = 0 

f'(x) = -(2x)/(x² + 2)² = 0

⇒ x = 0

At x = 0^–, f'(x) > 0

At x = 0⁺, f'(x) < 0

Therefore, local minimum and maximum value of f(0) = 1/2