EXERCISE SOLUTION-0.5, Calculus Book

Q.01-02 Simplify the expression without using a calculating utility. Solution: 1.  (a) $$\quad-8^{2 / 3}=-\left(2^3\right)^{2 / 3}=-2^2=-4$$ (b) $$\quad(-8)^{2 / 3}=\left[(-8)^2\right]^{1 / 3}=[16]^{1 / 3}=\left(4^3\right)^{1 / 3}=4$$ (c) $$8^{-2 / 3}=\left(2^3\right)^{-2 / 3}=(2)^{-2}=\frac{1}{2^2}=\frac{1}{4}$$ 2. (a) $$2^{-4}=\frac{1}{2^4}=\frac{1}{16}$$ (b) $$\quad 4^{1.5}=4^{15 / 10}=4^{3 / 2}=4^{3 / 2}=\left(2^2\right)^{3 / 2}=2^3=8$$ (c) $$9^{-0.5}=9^{-5 / 10}=\left(3^2\right)^{-1 / 2}=(3)^{-1}=\frac{1}{3}$$ Q.05 Find … Read more

EXERCISE SOLUTION-0.4, Pag#48, Calculus Book

Q.01 Determine whether $f$ and $g$ are inverse functions. Solution: (a) $$\begin{aligned} & f(x)=4 x, \quad g(x)=\frac{1}{4} x \\ \\ & f(g(x))=4 g(x)=4 \cdot \frac{1}{4} x=x \\ \\ & g(f(x))=\frac{1}{4} f(x)=\frac{1}{4} \cdot 4 x=x \\ \\ & \Rightarrow \quad f(g(x))=x, g(f(x))=x \end{aligned}$$ $\Rightarrow \quad f$ and $g$ are inverse functions. (b) $$\begin{aligned} f(x)=3 x+1, \quad … Read more

EXERCISE SOLUTION-0.1, Page#11, Calculus Book

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 … Read more

EXERCISE SOLUTION-0.2, Pag#24, Calculus Book

Q.32 Solution:  $$f(x)=\sqrt{x-3}, g(x)=\sqrt{x^{2}+3}$$ Domain of $f$ consists of all real numbers $x$ such that $x-3 \geq 0 \Rightarrow x \geq 3$ or $[3, \infty)$ Range of $f:[0, \infty)$ Domain of $g$ : All real numbers or $(-\infty, \infty)$ $$(f \circ g)(x)=f(g(x))$$ $$=f\left(\sqrt{x^{2}+3}\right)=\sqrt{\sqrt{x^{2}+3}-3}$$ $$(f \circ g)(x)=\sqrt{\sqrt{x^{2}+3}-3}$$ Domain of $f$ og: for all $x \geq \sqrt{6}$ … Read more