11类RD Sharma解决方案–第3章功能–练习3.4

问题1.在以下每一项中找到f + g，f – g，cf(c∈R，c≠0)，fg，1 / f和f / g：

(i)f(x)= x ³ +1和g(x)= x + 1

解决方案：

Given, f(x) = x³ + 1 and g(x) = x + 1 and f(x): R → R and g(x): R → R

We know, (f + g)(x) = f(x) + g(x) ⇒ (f + g) (x) = x³ + 1 + x + 1 = x³ + x + 2

So, (f + g)(x) = x³ + x + 2

As, (f – g)(x) = f(x) – g(x) ⇒ (f – g)(x) = x³ + 1 – (x + 1) = x³ + 1 – x – 1 = x³ –x

So, (f – g)(x) = x³ – x

Also, (cf)(x) = c × f(x) ⇒ (cf)(x) = c(x³ + 1) = cx³ + c

So, (cf)(x) = cx³ + c

Since, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = (x³ + 1)(x + 1) = (x + 1) (x² – x + 1) (x + 1) = (x + 1)² (x² – x + 1)

So, (fg)(x) = (x + 1)²(x² – x + 1)

Now, (1/f)(x) = 1/f (x) ⇒ 1/f (x) = 1/(x³ + 1)

Since 1/f(x) is undefined when f(x) = 0 or when x = – 1, 

Hence, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x³ + 1)  

Lastly, (f/g)(x) = f(x)/g(x) ⇒ (f/g) (x) = (x³ + 1)/(x + 1)

Since (x³ + 1)/(x + 1) is undefined when g(x) = 0 or when x = –1.

As x³ + 1 = (x + 1)(x² – x + 1), we have (f/g)(x) = $\frac{(x+1)(x^2-x+1)}{(x+1)}$= x² – x + 1

Hence, f/g: R – {–1} → R is given by (f/g)(x) = x² – x + 1

(ii)f(x)= $\sqrt{(x-1)}$和g(x)= $\sqrt{(x+1)}$

解决方案：

Since real square root is defined only for non-negative numbers, f(x): [1, ∞) → R+ and g(x): [–1, ∞) → R+

We know, (f + g)(x) = f(x) + g(x) = $\sqrt{(x-1)}$+ $\sqrt{(x+1)}$

Domain of (f + g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f + g: [1, ∞) → R is given by (f + g)(x) = $\sqrt{(x-1)}$+ $\sqrt{(x+1)}$

As, (f – g)(x) = f(x) – g(x) = $\sqrt{(x-1)}$– $\sqrt{(x+1)}$

Domain of (f – g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f – g: [1, ∞) → R is given by (f – g)(x) = $\sqrt{(x-1)}$– $\sqrt{(x+1)}$

Since, (cf)(x) = c × f(x) = c $\sqrt{(x-1)}$

Domain of (cf) = Domain of f = [1, ∞)

Hence, cf: [1, ∞) → R is given by (cf)(x) = c $\sqrt{(x-1)}$

Also, (fg)(x) = f(x)g(x) = $\sqrt{(x-1)}\sqrt{(x+1)}$= $\sqrt{x^2 -1}$

Domain of (fg) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, fg: [1, ∞) → R is given by (fg)(x) = $\sqrt{x^2 -1}$

Now, (1/f) (x) = 1/f(x) = $\frac{1}{\sqrt{x-1}}$

Domain of (1/f) = Domain of f = [1, ∞)

Since $\frac{1}{\sqrt{x-1}}$ is also undefined when x – 1 = 0 or x = 1.

Hence, 1/f: (1, ∞) → R is given by (1/f)(x) = $\frac{1}{\sqrt{x-1}}$

Lastly, (f/g) (x) = f(x)/g(x) =$\frac{\sqrt{x-1}}{\sqrt{x+1}}$

Domain of (f/g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f/g: [1, ∞) → R is given by (f/g) (x) =$\frac{\sqrt{x-1}}{\sqrt{x+1}}$

问题2。令f(x)= 2x + 5和g(x)= x ² + x。描述并查找每个域：

(i)f + g

解决方案：

Given, f(x) = 2x + 5 and g(x) = x² + x

Both f(x) and g(x) are defined for all x ∈ R.

So, domain of f = domain of g = R

We know, (f + g)(x) = f(x) + g(x) ⇒ (f + g)(x) = 2x + 5 + x² + x 

= x² + 3x + 5

Since (f + g)(x) Is defined for all real numbers x.

 Hence, (f + g)(x) = x² + 3x + 5 and the domain of (f + g) is R.

(ii)f – g

解决方案：

As (f – g)(x) = f(x) – g(x) ⇒ (f – g)(x) 

= 2x + 5 – (x² + x) 

= 2x + 5 – x² – x 

= 5 + x – x²

Since (f – g)(x) is defined for all real numbers x.

Hence, (f – g)(x) = 5 + x – x² and the domain of (f – g) is R.

(iii)fg

解决方案：

We know, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = (2x + 5)(x² + x) 

= 2x(x² + x) + 5(x² + x) 

= 2x³ + 2x² + 5x² + 5x 

= 2x³ + 7x² + 5x

Since (fg)(x) is defined for all real numbers x.

Hence, (fg)(x) = 2x³ + 7x² + 5x and the domain of fg is R.

(iv)f /克

解决方案：

We know, (f/g) (x) = f(x)/g(x) ⇒ (f/g)(x) = $\frac{2x+5}{x^2+x}$

Clearly (f/g)(x) is defined for all real values of x, except for the case when x² + x = 0.

x² + x = 0 ⇒ x(x + 1) = 0 ⇒ x = 0 or x + 1 = 0 ⇒ x = 0 or –1

When x = 0 or –1, (f/g)(x) will be undefined.

Hence, The domain of f/g = R – {–1, 0}

问题3.如果在[–2，2]上定义f(x)并由下式给出$\begin{cases}-1 \ \ \ ,-2\lex\le0\\x-1 \ \ \ ,0<x\le2\end{cases}$和g(x)= f(| x |)+ | f(x)|。求g(x)

解决方案：

f(|x|) = |x|  – 1, where –2 ≤ x ≤ 2

and, |f(x)| = $\begin{cases}1, \ \ -2\lex\le0\\-(x-1),\ \ 0\lex\le1\\(x-1), \ \ 1\lex\le2\end{cases}$

Thus, g(x) = f(|x|) + |f(x)| = $\begin{cases}-x, \ \ -2\lex\le0\\0, \ \ 0\lex\le1\\2, \ \ 1\lex\le2\end{cases}$

问题4.令f，g是由f(x)=定义的两个实函数$\sqrt{x+1}$和g(x)= $\sqrt{9-x^2}$ 。然后，描述以下每个功能。

(i)f + g

解决方案：

Given, f(x) = $\sqrt{x+1}$and g(x) = $\sqrt{9-x^2}$

f(x) takes real values only when x + 1 ≥ 0 x ≥ –1, x ∈ [–1, ∞) ⇒ Domain of f = [–1, ∞)

g(x) takes real values only when 9 – x² ≥ 0 

⇒ x² ≤ 9 ⇒ x² – 3² ≤ 0 ⇒ (x + 3)(x – 3) ≤ 0 

⇒ x ≥ –3 and x ≤ 3 

⇒ Domain of g = [–3, 3]

We know, (f + g)(x) = f(x) + g(x) ⇒ (f + g) (x) = $\sqrt{x+1}$ + $\sqrt{9-x^2}$

Domain of f + g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, f + g: [–1, 3] → R is given by (f + g)(x) = f(x) + g(x) = $\sqrt{x+1}$+ $\sqrt{9-x^2}$

(ii)g – f

解决方案：

We know, (g – f)(x) = g(x) – f(x) ⇒ (g – f)(x) =$\sqrt{9-x^2}$– $\sqrt{x+1}$

Domain of g – f = Domain of g ∩ Domain of f = [–3, 3] ∩ [–1, ∞) = [–1, 3]

Hence, g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = $\sqrt{9-x^2}$– $\sqrt{x+1}$

(iii)fg

解决方案：

We know, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = $\sqrt{x+1}\sqrt{9-x^2}$  

= $\sqrt{x(9-x^2) + (9-x^2)}$

= $\sqrt{9x-x^3+9-x^2}$ 

= $\sqrt{9+9x-x^2-x^3}$

Domain of fg = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = $\sqrt{x+1}\sqrt{9-x^2}$= $\sqrt{9+9x-x^2-x^3}$

(iv)f /克

解决方案：

We know, (f/g) (x) = f(x)/g(x) ⇒ (f/g) (x) = $\frac{\sqrt{x+1}}{\sqrt{9-x^2}}$ 

Domain of f/g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

However, (f/g) (x) is defined for all real values of x ∈ [–1, 3], except for the case when 9 – x2 = 0 or x = ± 3

When x = ±3, (f/g) (x) will be undefined as the division result will be indeterminate.

Domain of f/g = [–1, 3] – {–3, 3} = [–1, 3)

Hence, f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = $\frac{\sqrt{x+1}}{\sqrt{9-x^2}}$

(v)g / f

解决方案：

We know, (g/f)(x) = g(x)/f(x) ⇒ (g/f)(x) = $\frac{\sqrt{9-x^2}}{\sqrt{x+1}}$

Domain of g/f = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

However, (g/f) (x) is defined for all real values of x ∈ [–1, 3], except for the case when x + 1 = 0 or x = –1

When x = –1, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = [–1, 3] – {–1} = (–1, 3]

Hence, g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) =$\frac{\sqrt{9-x^2}}{\sqrt{x+1}}$  

(vi)2f –√5克

解决方案：

We know, (2f – √5g)(x) = 2f(x) – √5g(x) 

⇒ (2f – √5g)(x) = 2f (x) – √5g(x)

= 2 $\sqrt{x+1}$ – $\sqrt5\sqrt{9-x^2}$ 

= 2 $\sqrt{x+1}$ – $\sqrt{45- 5x^2}$

Domain of 2f – √5g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, 2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f(x) – √5 g(x) = 2 $\sqrt{x+1}$– $\sqrt{45- 5x^2}$

(vii)f ² + 7f

解决方案：

We know, (f² + 7f)(x) = f²(x) + (7f)(x) 

⇒ (f² + 7f) (x) = f(x).f(x) + 7f(x) 

=$\sqrt{x+1}.\sqrt{x+1}$ + 7 $\sqrt{x+1}$ 

= x + 1 + 7 $\sqrt{x+1}$

Domain of f² + 7f is same as domain of f = [–1, ∞)

Hence, f² + 7f: [–1, ∞) → R is given by (f2 + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7 $\sqrt{x+1}$

(viii)5 /克

解决方案：

We know, (5/g)(x) = 5/g(x) ⇒ (5/g)(x) = $\frac{5}{\sqrt{9-x^2}}$

Domain of 5/g = Domain of g = [–3, 3]

However, (5/g)(x) is defined for all real values of x ∈ [–3, 3], except for the case when 9 – x² = 0 or x = ± 3

When x = ±3, (5/g)(x) will be undefined.

Thus, Domain of 5/g = [–3, 3] – {–3, 3} = (–3, 3)

Hence, 5/g: (–3, 3) → R is given by (5/g)(x) = 5/g(x) = $\frac{5}{\sqrt{9-x^2}}$

问题5。如果f(x)= log _e (1 – x)并且g(x)= [x]，则确定以下每个函数：

(i)f + g

解决方案：

f(x) = log_e (1 – x) and g(x) = [x]

We know, f(x) takes real values only when 1 – x > 0 or when 1 > x

x < 1, ∴ x ∈ (–∞, 1) ⇒ Domain of f = (–∞, 1)

 g(x) is defined for all real numbers x. ⇒ Domain of g = [x], x ∈ R = R

Thus, (f + g)(x) = f(x) + g(x) = log_e(1 – x) + [x]

Domain of f + g = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

Hence, f + g: (–∞, 1) → R is given by (f + g) (x) = log_e(1 – x) + [x]

(ii)fg

解决方案：

We know, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = log_e(1 – x) × [x] = [x]log_e(1 – x)

Domain of fg = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

Hence, fg: (–∞, 1) → R is given by (fg) (x) = [x] log_e(1 – x).

(iii)f /克

解决方案：

We know, (f/g)(x) = f(x)/g(x) ⇒ (f/g)(x) = $\frac{log_e(1 - x)}{x}$

Domain of f/g = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

However, (f/g) (x) is defined for all real values of x ∈ (–∞, 1), except when [x] = 0.

We have, [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

When 0 ≤ x < 1, (f/g) (x) will be undefined. Domain of f/g = (–∞, 1) – [0, 1) = (–∞, 0)

Hence, f/g: (–∞, 0) → R is given by (f/g)(x) =$\frac{log_e(1 - x)}{x}$

(iv)g / f

解决方案：

We know, (g/f)(x) = g(x)/f(x) ⇒ (g/f)(x) = $\frac{x}{log_e (1 - x)}$

However, (g/f)(x) is defined for all real values of x ∈ (–∞, 1), except for the case when loge (1 – x) = 0. 

⇒ loge (1 – x) = 0 ⇒ 1 – x = 1 

or x = 0

When x = 0, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = (–∞, 1) – {0} = (–∞, 0) ∪ (0, 1)

Hence, g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f)(x) = $\frac{x}{log_e (1 - x)}$

(v)(f + g)(– 1)

解决方案：

(f + g) (x) = log_e(1 – x) + [x], x ∈ (–∞, 1)

Substituting x = –1 in the above equation, we get

(f + g)(–1) = log_e(1 – (–1)) + [–1] 

= log_e(1 + 1) + (–1) 

= log_e2 – 1

Hence, (f + g)(–1) = log_e2 – 1

(vi)(fg)(0)

解决方案：

We have, (fg)(x) = [x]log_e(1 – x), x ∈ (–∞, 1)

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]log_e(1 – 0) 

= 0 × log_e1 = 0

Hence, (fg) (0) = 0

(vii)(f / g)(1/2)

解决方案：

(f/g)(x) =  $\frac{log_e(1 - x)}{x}$, x ∈ (–∞, 0)

However, 1/2 is not in the domain of f/g.

Hence, (f/g)(1/2) does not exist.

(viii)(g / f)(1/2)

解决方案：

We have, (g/f)(x) =$\frac{x}{log_e (1 - x)}$, x ∈ (–∞, 0) ∪ (0, ∞)

Substituting x=1/2 in the above equation, we get

(g/f)(1/2) = $\frac{x}{log(1 - x)}$

= $\frac{\frac{1}{2}}{log_e(1 -\frac{1}{2})}$ 

= $\frac{0.5}{log_e(\frac{1}{2})}$ 

=$\frac{0}{loge(1/2)}$ = 0

Hence, (g/f)(1/2) = 0 

问题6.如果f，g，h是由f(x)=定义的实函数$\sqrt{x+1}$ ，g(x)= 1 / x和h(x)= 2x ² – 3，然后找到(2f + g – h)(1)和(2f + g – h)(0)的值。

解决方案：

Since, f(x) is defined for x + 1 ≥ 0 ⇒ x ≥ – 1 

 ⇒ x ∈ [– 1, ∞] = Domain of f(x)

Now, (2f+g –h)(x) = 2f(x) + g(x) – h(x) 

= 2 $\sqrt{x+1}$+ 1/x  – (2x² – 3)

 = 2 $\sqrt{1+1}$+ 1/1 – 2(1)² + 3

(2f + g – h)(x) = 2√2 + 2

Since (2f+g –h)(0) does not lie in the domain x ∈ [– 1, ∞] – 0, it does not exist.

问题7.函数f(x)的定义为：f(x)= $\begin{cases}1-x , \ \ x<0\\1 , \ \ x=0\\1+x , \ \ x>0\end{cases}$ 。绘制f(x)的图。

解决方案：

The graph of f(x) for x < 0 lies to the left of origin.

The graph of f(x) for x > 0 lies to the right of origin.

The graph of f(x) for x = 0 is represented by the point (0,1).

问题8.设f，g：R⇒R分别定义为f(x)= x +1和g(x)= 2x –3。求f + g，f – g和f / g。

解决方案：

We know, (f + g)(x) = f(x) + g(x) =  (x + 1 + 2x – 3) = 3x – 2.

Now, (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = 4 – x.

f/g(x) = f(x)/g(x) = $\frac{x+1}{2x-3}$

问题9.令f和g分别定义为f(x)=√x和g(x)= x。找出f + g，f – g，fg和f / g。

解决方案：

(f + g)(x) = f(x) + g(x) = √x + x 

(f – g)(x) = f(x) – g(x) = √x – x

(fg)(x) = f(x)g(x) = √x.x = x^{1/2 + 1} = x^3/2

f/g(x) = f(x)/g(x) = √x/x = 1/√x 

问题10.令f(x)= x ²和g(x)= 2x + 1是两个实函数。找出f + g，f – g，fg和f / g。

解决方案：

(f + g)(x) = f(x) + g(x) = x² + 2x + 1 = (x + 1)²

(f – g)(x) = f(x) – g(x) = x² – 2x – 1

(fg)(x) = f(x)g(x) = x²(2x + 1) = 2x³ + x²

f/g(x) = f(x)/g(x) = $\frac{x^2}{2x+1}$