Question 1: Find f+g, f-g, cf (c \in R, c \neq 0), fg, \frac{1}{f} and \frac{f}{g}  in each of the following:

i) f(x) = x^3 + 1 and g(x) = x + 1      ii) f(x) = \sqrt{x-1} and g(x) = \sqrt{x+1}

Answer:

i) Given f(x) = x^3 + 1 and g(x) = x + 1

f+g : R \rightarrow R is given by

(f+g)(x) = f(x) + g(x) = x^3 + 1 + x + 1 = x^3 + x + 2

 f-g : R \rightarrow R is given by

(f-g)(x) = f(x) - g(x) = x^3 + 1 - x - 1 = x^3 - x

cf : R \rightarrow R is given by

(cf)(x) = c f(x) = c(x^3 + 1)

fg : R \rightarrow R is given by

(fg)(x) = f(x).g(x) = (x^3+1)(x+1) = x^4 + x + x^3 + 1

\frac{1}{f} : R \rightarrow R is given by

\Big( \frac{1}{f} \Big) (x) = \frac{1}{f(x)} = \frac{1}{x^3+1}

\frac{f}{g} : R \rightarrow R is given by

\Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{x^3+1}{x+1} = \frac{(x+1)(x^2-x+1)}{x+1} = x^2-x+1

ii) Given f(x) = \sqrt{x-1} and g(x) = \sqrt{x+1}

f+g : [1, \infty ) \rightarrow R is given by

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

 f-g : [1, \infty ) \rightarrow R is given by

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

cf : [1, \infty ) \rightarrow R is given by

(cf)(x) = c f(x) = c(\sqrt{x-1})

fg : [1, \infty ) \rightarrow R is given by

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

\frac{1}{f} : [1, \infty ) \rightarrow R is given by

\Big( \frac{1}{f} \Big) (x) = \frac{1}{f(x)} = \frac{1}{\sqrt{x-1}}

\frac{f}{g} : [1, \infty ) \rightarrow R is given by

\Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x-1}}{\sqrt{x+1}}

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Question 2: Let f(x) = 2x+ 5 and g(x) = x^2 + x . Describe:

i) f+g    ii) f-g      iii) fg    iv) \frac{f}{g} . Find the domain in each case.

Answer:

We observe f(x) = 2x+5 is defined for all x \in R . Hence Domain (f) = R

Also g(x) = x^2 + x is defined for all x \in R . Hence Domain (g) = R

Therefore Domain (f) \cap Domain (g) = R

i)  f+g : R \rightarrow R is given by

(f+g)(x) = f(x) + g(x) = 2x+5 + x^2 + x = x^2 + 3x + 5

Hence Domain (f+g) = R

ii)  f-g : R \rightarrow R is given by

(f-g)(x) = f(x) - g(x) = 2x+5 - x^2 - x = x + 5 -x^2

Hence Domain (f-g) = R

iii)fg : R \rightarrow R is given by

(fg)(x) = f(x).g(x) = (2x+5)(x^2+x) = 2x^3+7x^2+5x

Hence Domain (fg) = R

iv) \frac{f}{g} : R \rightarrow R is given by

\Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{2x+5}{x^2+x}

Now for x^2 + x \neq 0 for all x except x = 0, -1

Hence Domain \Big( \frac{f}{g} \Big)  = R - \{ 0, -1 \}

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Question 3: If f(x)   be defined on [-2, 2] and is given by 

f(x) = \Bigg\{  \begin{array}{ll} -1, \hspace*{1.0cm} -2 \leq x \leq 0 \\ x-1, \hspace*{0.8cm} 0 < x \leq 2   \end{array}       and    g(x) = f(|x|) + |f(x)| . Find g(x) .

Answer:

Given f(x) = \Bigg\{  \begin{array}{ll} -1, \hspace*{1.0cm} -2 \leq x \leq 0 \\ x-1, \hspace*{0.8cm} 0 < x \leq 2   \end{array}

Now for g(x) = f(|x|) + g(|x|)

= \Bigg\{  \begin{array}{lll} x-1+1, \hspace*{1.8cm} -2 \leq x \leq 0 \\ x-1+(-x+1), \hspace*{0.8cm} 0 < x < 1 \\ x-1+x-1, \hspace*{1.3cm} 1 \leq x \leq 2   \end{array}

= \Bigg\{  \begin{array}{lll} x, \hspace*{1.8cm} -2 \leq x \leq 0 \\ 0, \hspace*{0.8cm} 0 < x < 1 \\ 2x-2, \hspace*{1.3cm} 1 \leq x \leq 2   \end{array}

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Question 4: Let f and g be two real functions defined by f(x) = \sqrt{x+1} and g(x) = \sqrt{9-x^2} . Then, describe each of the following functions:

i) f+g    ii) g-f    iii) fg    iv) \frac{f}{g}      v) \frac{g}{f}      vi) 2f - \sqrt{5} g    vii) f^2 + 7f     viii) \frac{5}{g}

Answer:

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

We observe that f(x) = \sqrt{x+1} is defined for x \geq -1

Hence Domain (f) = [-1, \infty)

Similarly, g(x) = \sqrt{9-x^2} is defined for 9-x^2 \geq 0 \Rightarrow x^2 - 9 \leq 0

\Rightarrow (x-3)(x+3) \leq 0

\Rightarrow x \in [-3, 3 ]

Hence Domain (g) = [-3, 3]

Now Domain (f) \cap Domain (g) = [-1, \infty) \cap [-3, 3] = [-1, 3]

i) f+g :[-1, 3] \rightarrow R is given by

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

ii) g-f : [-1, 3] \rightarrow R is given by

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

iii) (fg) : [-1, 3] \rightarrow R is given by

(fg)(x) = f(x) g(x) = \sqrt{x+1} \sqrt{9-x^2}

iv) \frac{f}{g} : [-1, 3 ) \rightarrow R is given by

\Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+1}}{\sqrt{9-x^2}}

v) \frac{g}{f} : [-1, 3 ) \rightarrow R is given by

\Big( \frac{g}{f} \Big) (x) = \frac{g(x)}{f(x)} = \frac{\sqrt{9-x^2}}{\sqrt{x+1}}

vi) 2f-\sqrt{5} g : [-1, 3) \rightarrow R is defined by

(2f-\sqrt{5} g)(x) = 2f(x) - \sqrt{5} g(x) = 2 \sqrt{x+1} - \sqrt{5} (\sqrt{9-x^2})

= 2 \sqrt{x+1} - \sqrt{45-5x^2}

vii) f^2 + 7f : [-1, \infty) \rightarrow R is given by

(f^2 + 7f)(x) = [f(x)]^2 + 7f(x) = (\sqrt{x+1})^2 + 7 \sqrt{x+1} = x+1 + 7 \sqrt{x+1}

viii) \frac{5}{g(x)} = \frac{5}{\sqrt{9-x^2}}  will be real for all values of x except for x \geq 3 and x \leq -3

Therefore \frac{5}{g} : (-3, 3) \rightarrow R is defined by

\Big( \frac{5}{g} \Big) (x)  = \frac{5}{\sqrt{9-x^2}}

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Question 5: If f(x) = \log_e (1-x) and g(x) = [x] , then determine each of the following functions:

i) f+g    ii) fg     iii) g-f     iv) \frac{f}{g}      v) \frac{g}{f}  

Also find v) (f+g)(-1), \ \ \ (fg) (0), \ \ \  \ \ \ \Big( \frac{f}{g} \Big) \Big( \frac{1}{2} \Big), \ \ \ \Big( \frac{g}{f} \Big) \Big( \frac{1}{2} \Big)

Answer:

Given f(x) = \log_e (1-x) and g(x) = [x]

For f(x) = \log_e (1-x) is defined for all (1-x) > 0 \Rightarrow x < 1 \Rightarrow x \in (-\infty, 1)

Therefore Domain (f) = (-\infty, 1)

For g(x) = [x] is defined for all x \in R

Therefore Domain (g) = R

Therefore Domain (f) \cap Domain (g) = (-\infty, 1) \cap R = (-\infty, 1)

i) f+g :(-\infty, 1) \rightarrow R is given by

(f+g)(x) = f(x) + g(x) = \log_e (1-x) +  g(x) = [x]

ii) (fg) : (-\infty, 1) \rightarrow R is given by

(fg)(x) = f(x) g(x) = (\log_e (1-x) ).(  g(x) ) = [x]

iii) Domain (f) = (-\infty, 1)

\frac{1}{g} = \frac{1}{[x]} will be real for all x except [x] = 0

\Rightarrow x \notin (0, 1)

Therefore Domain \Big( \frac{f}{g} \Big) = Domain (f) \cap Domain ( \frac{1}{g} )

=  (-\infty, 1) \cap R - \notin (0, 1) = (-\infty, 0)

\therefore \Big( \frac{f}{g} \Big) : (-\infty, 0) \rightarrow R is defined as 

\Big( \frac{f}{g} \Big)(x) = \frac{f(x)}{g(x)} = \frac{\log_e (1-x)}{[x] }

iv)  \frac{1}{f(x)} = \frac{1}{\log_e (1-x)}

\therefore \frac{1}{f(x)} is real if \log_e (1-x) \neq 0

\Rightarrow 10 x > 0 and 1-x \neq 0

\Rightarrow x < 1 and x \neq 1

\Rightarrow x \in (-\infty, 0) \cup (0, 1)

\therefore Domain ( \frac{g}{f} ) = Domain (g) \cap Domain ( \frac{1}{f} )

= R \cap (-\infty, 0) \cup (0, 1) = (-\infty, 0) \cup (0, 1)

\therefore \frac{g}{f} : (-\infty, 0) \cup (0, 1) is defined by

\Big( \frac{g}{f} \Big)(x) = \frac{g(x)}{f(x)} = \frac{[x] }{\log_e (1-x)}

v)    (f+g) (-1) = f(-1) + f(-1) = \log_e (1-(-1)) + [-1] = \log_e 2 - 1

(fg)(0) = f(0)g(0) = \log_e (1-0) + [0] = 0

\Big( \frac{f}{g} \Big) \Big( \frac{1}{2} \Big)  = \Big( \frac{f(\frac{1}{2})}{g(\frac{1}{2})} \Big) = \frac{\log_e (1-\frac{1}{2})}{[\frac{1}{2}]} =  does not exist because [\frac{1}{2}] = 0

\Big( \frac{g}{f} \Big) \Big( \frac{1}{2} \Big)  = \Big( \frac{g(\frac{1}{2})}{f(\frac{1}{2})} \Big) = \frac{[\frac{1}{2}]}{\log_e (1-\frac{1}{2})} = 0

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Question 6: If f, g, h are real functions defined by f(x) = \sqrt{x+1}, \ g(x) = \frac{1}{x} and h(x) = 2x^2 - 3 , then find the values of (2f + g- h)(1) and (2f+g-h)(0) .

Answer:

Given f(x) = \sqrt{x+1}, \ g(x) = \frac{1}{x} and h(x) = 2x^2 - 3

f(x) will be real for all x \geq -1 \Rightarrow Domain (f) = [-1, \infty)

g(x) will be real for all x except x = 0 \Rightarrow Domain (g) = R - \{0 \}

h(x) will be real for all x \in R \Rightarrow Domain (h) = R

\therefore Domain (f) \cap Domain (g) \cap Domain (h) = [-1, \infty) \cap R - \{0 \} \cap   R = [-1, \infty) - \{0 \}

\therefore 2f+g - h : [-1, \infty) - \{0 \} \rightarrow R is given by

(2f+g - h)(x) = 2f(x) + g(x) - h(x) = 2\sqrt{x+1} + \frac{1}{x} - 2x^2 - 3

(2f+g - h)(1) = 2f(1) + g(1) - h(1) = 2\sqrt{1+1} + \frac{1}{1} - 2(1)^2 - 3 = 2\sqrt{2}+2

(2f+g - h)(0) does not exist as x does not lie in domain [-1, \infty) - \{0 \}

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Question 7: The function f is defined by 

f(x) = \Bigg\{  \begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0  \\ x+1, \hspace*{1.0cm} x > 0   \end{array}

Draw the graph of f(x)

Answer:

Given f(x) = \Bigg\{  \begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0  \\ x+1, \hspace*{1.0cm} x > 0   \end{array}

2019-09-17_12-26-12

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Question 8: Let f, \ g \colon R \rightarrow R be defined respectively by f(x) = x + 1, \ g(x) = 2x-3 . Find f+g, \ f-g , and \frac{f}{g} .

Answer:

Given f(x) = x + 1, \ g(x) = 2x-3

f(x) will be real for all x \in R \Rightarrow Domain (f) = R

g(x ) will be real for all x \in R \Rightarrow Domain (g) = R

\therefore Domain (f) \cap Domain (g) = R

f+g  : R \rightarrow R is defined by  (f+g)(x)  = f(x) + g(x) = 3x-2

f-g  : R \rightarrow R is defined by  (f-g)(x)  = f(x) - g(x) = -x+4

\frac{1}{g} = \frac{1}{2x-3}  is real for 2x-3 \neq 0 \Rightarrow x \neq \frac{3}{2}

\frac{f}{g} : R - \{ \frac{3}{2} \} \rightarrow R   is give by \Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{x+1}{2x-3}

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Question 9: Let f \colon [ 0, \infty ) \rightarrow R and g \colon R \rightarrow R be defined by f(x) = \sqrt{x} and g(x) = x . Find f+g, \ f-g, fg and \frac{f}{g} .

Answer:

Given f \colon [ 0, \infty ) \rightarrow R and f(x) = \sqrt{x}

g \colon R \rightarrow R and g(x) = x

Domain (f) \cap Domain (g) = [0, \infty)

\therefore f+g : [0, \infty) \rightarrow R is defined as (f+g)(x) = f(x) + g(x)  = \sqrt{x}+x

f-g : [0, \infty) \rightarrow R is defined as (f-g)(x) = f(x) - g(x)  = \sqrt{x}-x

\frac{f}{g} : (0, \infty) \rightarrow R is defined as  \Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}

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Question 10: Let f(x) = x^2 and g(x) = 2x+ 1 be two real functions. Find (f+g)(x), \ (f-g)(x), \ (fg)(x) and \Big( \frac{f}{g} \Big) (x)

Answer:

Given f(x) = x^2 and g(x) = 2x+ 1

f(x) will be real for all x \in R \Rightarrow Domain (f) = R

Similarly, g(x) will be real for all x \in R \Rightarrow Domain (g) = R

\therefore f+g : R \rightarrow R is defined as (f+g)(x) = f(x) + g(x)  = x^2+2x+ 1

\therefore f-g : R \rightarrow R is defined as (f-g)(x) = f(x) - g(x)  = x^2-2x- 1

\therefore fg : R \rightarrow R is defined as (fg)(x) = f(x).g(x)  = x^2(2x+ 1)

\frac{f}{g} : R - \{ \frac{1}{2} \} \rightarrow R is defined as  \Big( \frac{f}{g} \Big) (x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x+1}

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