Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/33

Rh In each case $$\scriptstyle{y}$$ (the value of the function) is known, or, as we say, defined, for all values of $$\scriptstyle{x}$$. This is not by any means true of all functions, as the following examples illustrating the more common exceptions will show.

(1)&emsp;$$\scriptstyle{y=\frac{a}{x-b}}$$.

Here the value of $$\scriptstyle{y}$$ (i.e. the function) is defined for all values of $$\scriptstyle{x}$$ except $$\scriptstyle{x=b}$$. When $$\scriptstyle{x=b}$$ the divisor becomes zero and the value of $$\scriptstyle{y}$$ cannot be computed from (1). Any value might be assigned to the function for this value of the argument.

(2)&emsp;$$\scriptstyle{y=\sqrt{x}}$$.

In this case the function is defined only for positive values of $$\scriptstyle{x}$$. Negative values of $$\scriptstyle{x}$$ give imaginary values for $$\scriptstyle{y}$$, and these must be excluded here, where we are confining ourselves to real numbers only.

(3)&emsp;$$\scriptstyle{y=\log_a x.\qquad\qquad\qquad a>0}$$

Here $$\scriptstyle{y}$$ is defined only for positive values of $$\scriptstyle{x}$$. For negative values of $$\scriptstyle{x}$$ this function does not exist (see § 19).

(4)&emsp;$$\scriptstyle{y=\operatorname{arc~sin}x, y=\operatorname{arc~cos}x}$$.

Since sines and cosines cannot become greater than $$\scriptstyle{+1}$$ nor less than $$\scriptstyle{-1}$$, it follows that the above functions are defined for all values of $$\scriptstyle{x}$$ ranging from $$\scriptstyle{-1}$$ to $$\scriptstyle{+1}$$ inclusive, but for no other values.

1. Given $$\scriptstyle{f(x)=x^3-10x^2+31x-30}$$; show that

2. If $$\scriptstyle{f(x)=x^3-3x+2}$$, find $$\scriptstyle{f(0),\ f(1),\ f(-1),\ f(-\frac{1}{2}),\ f(1\frac{1}{3})}$$.

3. If $$\scriptstyle{f(x)=x^3-10x^2+31x-30}$$, and $$\scriptstyle{\phi(x)=x^4-55x^2-210x-216}$$, show that

4. If $$\scriptstyle{F(x)=2^x}$$, find $$\scriptstyle{F(0),\ F(-3),\ F(\frac{1}{2}),\ F(-1)}$$.

5. Given $$\scriptstyle{F(x)=x(x-1)(x+6)(x-\frac{1}{2})(x+\frac{5}{4})}$$; show that