EXERCISE SOLUTION-0.1, Page#11, Calculus Book

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EXERCISE SOLUTION-0.1, Page#11, Calculus Book [10th Edition] by Howard ANTON, IRL BEVINS, and STEPHAN DAVIS. This section of the book consists of the definition of functions and their domain and range.  The functions can be described by tables, graphs, formulas, and words.  You can find the solutions to problems in this section which will help to get the concept of functions. 

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Q.02
Solution:
(a) For x=-1, and x=4, y=1.
(b) There is no value of x for which y=3.
(c) For y=-1, x=3.
(d) For x=0,3,5, y \leqslant 0.
(e) maximum value of y=9 at x=6 miniment value of y=-2 at x=0.
Q.03
Solution:
(a) It is a function because no vertical line intersects the graph more than once.
(b) It is a function because no vertical line intersects the graph more than once.
(c) It is not a function because the vertical line test fails.
(d) It is not a function because the vertical line test fails.
 
Q.04
Solution:

(a)

    \[f(x)=\frac{x^2+1}{x+1} ; \quad g(x)=x\]


The function f is not defined at x=-1 Therefore, the natural domain of f consists of all real numbers except x=-1.

The natural domain of g(x) consists of all real numbers ie (-\infty,+\infty).

Natural domain of f=(-\infty,-1) \cup(-1, \infty)
Natural domain of g=(-\infty, \infty)
Natural domain of f= natural domain g if x=-1 is excluded from the natural of g. i.e. f(x)=g(x) on (-\infty,-1) \cup(-1, \infty).
(b)

    \[\quad f(x)=\frac{x \sqrt{x}+\sqrt{x}}{x+1}: \quad g(x)=\sqrt{x}\]


The function f has real values for all x \geq 0.
\Rightarrow Natural domain of f(x)=x \geq 0 or [0,+\infty)
The function g has real values for all x \geq 0.
\Rightarrow Natural domain of g(x)=x \geqslant 0 or [0,+\infty).
Therefore the natural domain of f= Natural domain of g

 
Q.07
Solution:

(a)

    \[f(x)=3 x^2-2\]


    \[\begin{aligned}& f(0)=3(0)^2-2=-2 \\ \\& f(2)=3(+2)^2-2=10 \\ \\& f(-2)=3(-2)^2-2=10 \\ \\& f(3)=3(3)^2-2=25 \\ \\& f(\sqrt{2})=3(\sqrt{2})^2-2=4 \\ \\& f(3 t)=3(3 t)^2-2=27 t^2-2\end{aligned}\]

Q.07b
Solution:

    \[\begin{aligned}& f(x)= \begin{cases}\frac{1}{x}, & x>3 \\ \\2 x & x \leq 3\end{cases} \\ \\& f(0)=2(0)=0 \\ \\& f(2)=2(2)=4 \\ \\& f(-2)=2(-2)=-4 \\ \\& f(3)=2(3)=6 \\ \\& f(\sqrt{2})=2 \sqrt{2}, \\ \\& f(3 t)=\left\{\begin{array}{l}\frac{1}{3 t} \quad 3 t>3\\ \\2(3 t) \quad 3 t \leq 3\end{array}\right. \\ \\& f(3 t)= \begin{cases}\frac{1}{3 t} & t > 1\\ \\6 t & t \leq 1\end{cases}\end{aligned}\]

Q.08a
Solution:

    \[\quad g(x)=\frac{x+1}{x-1}\]


    \[g\left(t^2-1\right)=\frac{\left(t^2-1\right)+1}{\left(t^2-1\right)-1}=\frac{t^2-1+1}{t^2-1-1}=\frac{t^2}{t^2-2}\]

Q.08b
Solution:

    \[\begin{aligned}g(x) & = \begin{cases}\sqrt{x+1} & x \geqslant 1 \\ \\3 & x<1\end{cases} \\ \\g\left(t^2-1\right) & = \begin{cases}\sqrt{t^2-1+1} & t^2-1 \geqslant 1 \\ \\3 & t^2-1<1\end{cases}\end{aligned}\]

    \[\begin{aligned}& g\left(t^2-1\right)= \begin{cases}\sqrt{t^2} & t^2 \geqslant 1+1 \\ \\3 & t^2<1+1\end{cases} \\ \\& g\left(t^2-1\right)= \begin{cases}|t|, & t^2 \geqslant 2 \\ \\3 & t^2<2\end{cases}\end{aligned}\]

Q.09 Find the natural domain and determine the range of each function.
Solution:
(d)

    \[\begin{aligned}& G(x)=\sqrt{x^2-2 x+5} \\ \\& x^2-2 x+5=x^2-2 x+1+4 \\ \\& =(x-1)^2+4 \geqslant 4 \\ \\& G(x)=\sqrt{x^2-2 x+5}=\sqrt{(x-1)^2+4} \geqslant \sqrt{4}=2 \\ \\& \Rightarrow G(x) \geq 2 \quad \forall x \\ \\&\end{aligned}\]


\therefore G(x) is defined for all x.
Range =\quad y \geqslant 2

Q.09(e)
Solution:

    \[</div> <div> \begin{aligned}& h(x)=\frac{1}{1-\sin x} \\ \\& h(x) \text { is defined for those} x \text { for which } 1-\sin x \neq 0 \\ \\ & \sin x \neq 1 \Rightarrow x \neq \sin ^{-1}(1) \\ \\ & \Rightarrow x \neq \text { all odd multiple of } \frac{\pi}{2} \\ \\ & \Rightarrow \quad x \neq\left(2 n+\frac{1}{2}\right) \pi, \quad n=0, \pm 1 \pm 2, \ldots \\ \\& \Rightarrow \quad -1\leq \sin x < 1 \quad \forall \quad x \neq\left(2 n+\frac{1}{2}\right) \pi, \quad n=0, \pm 1 \pm 2, \ldots\\ \end{aligned}\]

Multiplying by -1

    \[\begin{aligned} & \Rightarrow \quad -1<-\sin x \leq 1 \\ \\ & \Rightarrow \quad -1+1<-\sin x+1 \leq 1+1 \\ \\ & \Rightarrow \quad 0<-\sin x+1 \leq 2 \\ \\ & \Rightarrow \quad -\sin x+1 \leq 2 \\ \\ & \Rightarrow \quad \frac{1}{1-\sin x} \geq \frac{1}{2} \\ \\& \Rightarrow \quad h(x) \geqslant \frac{1}{2} \\ \\& \therefore \quad \text { Range: } y \geq \frac{1}{2} .\end{aligned}\]

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