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

$\displaystyle \text{(i) } f(x) = x^3 + 1 \text{ and } g(x) = x + 1 \hspace{0.5cm} \text{(ii) } f(x) = \sqrt{x-1} \text{ and } g(x) = \sqrt{x+1}$

$\displaystyle \text{(i) Given } f(x) = x^3 + 1 \text{ and } g(x) = x + 1$

$\displaystyle f+g : R \rightarrow R \text{ is given by }$

$\displaystyle (f+g)(x) = f(x) + g(x) = x^3 + 1 + x + 1 = x^3 + x + 2$

$\displaystyle f-g : R \rightarrow R \text{ is given by }$

$\displaystyle (f-g)(x) = f(x) - g(x) = x^3 + 1 - x - 1 = x^3 - x$

$\displaystyle cf : R \rightarrow R \text{ is given by }$

$\displaystyle (cf)(x) = c f(x) = c(x^3 + 1)$

$\displaystyle fg : R \rightarrow R \text{ is given by }$

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

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

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

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

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

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

$\displaystyle f+g : [1, \infty ) \rightarrow R \text{ is given by }$

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

$\displaystyle f-g : [1, \infty ) \rightarrow R \text{ is given by }$

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

$\displaystyle cf : [1, \infty ) \rightarrow R \text{ is given by }$

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

$\displaystyle fg : [1, \infty ) \rightarrow R \text{ is given by }$

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

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

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

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

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

$\displaystyle \\$

$\displaystyle \text{Question 2: Let } f(x) = 2x+ 5 \text{ and } g(x) = x^2 + x \text{. Describe:}$

$\displaystyle \text{(i) } f+g \hspace{1.0cm} \text{(ii) } f-g \hspace{1.0cm} \text{(iii) } fg \hspace{1.0cm} \text{(iv) } \frac{f}{g} \text{ Find the domain in each case. }$

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

Also $\displaystyle g(x) = x^2 + x$ is defined for all $\displaystyle x \in R$. $\displaystyle \text{Hence Domain } (g) = R$

Therefore Domain $\displaystyle (f) \cap \text{Domain } (g) = R$

$\displaystyle \text{(i) } f+g : R \rightarrow R \text{ is given by }$

$\displaystyle (f+g)(x) = f(x) + g(x) = 2x+5 + x^2 + x = x^2 + 3x + 5$

$\displaystyle \text{Hence Domain } (f+g) = R$

$\displaystyle \text{(ii) } f-g : R \rightarrow R \text{ is given by }$

$\displaystyle (f-g)(x) = f(x) - g(x) = 2x+5 - x^2 - x = x + 5 -x^2$

$\displaystyle \text{Hence Domain } (f-g) = R$

$\displaystyle \text{(iii) } fg : R \rightarrow R \text{ is given by }$

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

$\displaystyle \text{Hence Domain } (fg) = R$

$\displaystyle \text{(iv) } \frac{f}{g} : R \rightarrow R \text{ is given by }$

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

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

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

$\displaystyle \\$

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

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

$\displaystyle \text{Find } g(x).$

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

Now for $\displaystyle g(x) = f(|x|) + g(|x|)$

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) } f+g \hspace{1.0cm} \text{(ii) } g-f \hspace{1.0cm} \text{(iii) } fg \hspace{1.0cm} \text{(iv) } \frac{f}{g} \hspace{1.0cm} \text{(v) } \frac{g}{f} \hspace{1.0cm} \\ \\ \text{(vi) } 2f - \sqrt{5} g \hspace{1.0cm} \text{(vii) } f^2 + 7f \hspace{1.0cm} \text{(viii) } \frac{5}{g}$

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

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

$\displaystyle \text{Hence Domain } (f) = [-1, \infty)$

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

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

$\displaystyle \Rightarrow x \in [-3, 3 ]$

$\displaystyle \text{Hence Domain } (g) = [-3, 3]$

Now Domain $\displaystyle (f) \cap \text{Domain } (g) = [-1, \infty) \cap [-3, 3] = [-1, 3]$

$\displaystyle \text{(i) } f+g :[-1, 3] \rightarrow R \text{ is given by }$

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

$\displaystyle \text{(ii) } g-f : [-1, 3] \rightarrow R \text{ is given by }$

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

$\displaystyle \text{(iii) } (fg) : [-1, 3] \rightarrow R \text{ is given by }$

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

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

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

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

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

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

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

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

v$\displaystyle \text{(ii) } f^2 + 7f : [-1, \infty) \rightarrow R \text{ is given by }$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) } f+g \hspace{1.0cm} \text{(ii) } fg \hspace{1.0cm} \text{(iii) } g-f \hspace{1.0cm} \text{(iv) } \frac{f}{g} \hspace{1.0cm} \text{(v) } \frac{g}{f} \hspace{1.0cm} \\ \\ \text{(v) } (f+g)(-1), (fg) (0), \Big( \frac{f}{g} \Big) \Big( \frac{1}{2} \Big), \Big( \frac{g}{f} \Big) \Big( \frac{1}{2} \Big)$

$\displaystyle \text{Given } f(x) = \log_e (1-x) \text{ and } g(x) = [x]$

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

$\displaystyle \text{Therefore Domain } (f) = (-\infty, 1)$

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

$\displaystyle \text{Therefore Domain } (g) = R$

$\displaystyle \text{Therefore Domain } (f) \cap \text{Domain } (g) = (-\infty, 1) \cap R = (-\infty, 1)$

$\displaystyle \text{(i) } f+g :(-\infty, 1) \rightarrow R \text{ is given by }$

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

$\displaystyle \text{(ii) } (fg) : (-\infty, 1) \rightarrow R \text{ is given by }$

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

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

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

$\displaystyle \Rightarrow x \notin (0, 1)$

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

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

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

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

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

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

$\displaystyle \Rightarrow 10 x > 0 \text{ and } 1-x \neq 0$

$\displaystyle \Rightarrow x < 1 \text{ and } x \neq 1$

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

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

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

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

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

$\displaystyle \text{(v) } (f+g) (-1) = f(-1) + f(-1) = \log_e (1-(-1)) + [-1] = \log_e 2 - 1$

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

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

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

$\displaystyle \\$

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

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

$\displaystyle f(x)$ will be real for all $\displaystyle x \geq -1 \Rightarrow \text{Domain } (f) = [-1, \infty)$

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

$\displaystyle h(x)$ will be real for all $\displaystyle x \in R \Rightarrow \text{Domain } (h) = R$

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

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

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

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

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

$\displaystyle \\$

Question 7: The function f is defined by

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

Draw the graph of $\displaystyle f(x)$

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

$\displaystyle \\$

Question 8: Let $\displaystyle f, g \colon R \rightarrow R$ be defined respectively by $\displaystyle f(x) = x + 1, g(x) = 2x-3.$

$\displaystyle \text{Find } f+g, f-g \text{, and } \frac{f}{g}.$

$\displaystyle \text{Given } f(x) = x + 1, g(x) = 2x-3$

$\displaystyle f(x)$ will be real for all $\displaystyle x \in R \Rightarrow \text{Domain } (f) = R$

$\displaystyle g(x$) will be real for all $\displaystyle x \in R \Rightarrow \text{Domain } (g) = R$

$\displaystyle \therefore \text{Domain } (f) \cap \text{Domain } (g) = R$

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

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

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

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

$\displaystyle \\$

Question 9: Let $\displaystyle f \colon [ 0, \infty ) \rightarrow R \text{ and } g \colon R \rightarrow R$ be defined by $\displaystyle f(x) = \sqrt{x} \text{ and } g(x) = x.$

$\displaystyle \text{Find } f+g, f-g, fg \text{ and } \frac{f}{g}.$

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

$\displaystyle g \colon R \rightarrow R \text{ and } g(x) = x$

Domain $\displaystyle \text{ } (f) \cap \text{Domain } (g) = [0, \infty)$

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{Given } f(x) = x^2 \text{ and } g(x) = 2x+ 1$

$\displaystyle f(x)$ will be real for all $\displaystyle x \in R \Rightarrow \text{Domain } (f) = R$

Similarly, $\displaystyle g(x)$ will be real for all $\displaystyle x \in R \Rightarrow \text{Domain } (g) = R$

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

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

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

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