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

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) $\scriptstyle {y={\frac {a}{x-b}}}$.

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).^[1] Any value might be assigned to the function for this value of the argument.

(2) $\scriptstyle {y={\sqrt {x}}}$.

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) $\scriptstyle {y=\log _{a}x.\qquad \qquad \qquad a>0}$

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) $\scriptstyle {y=\operatorname {arc~sin} x,y=\operatorname {arc~cos} x}$.

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.

 

EXAMPLES

 

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

${\begin{array}{c c}{\begin{aligned}\scriptstyle {f(0)}&\scriptstyle {=-30,}\\\scriptstyle {f(2)}&\scriptstyle {=0,}\\\scriptstyle {f(3)}&\scriptstyle {=f(5),}\\\scriptstyle {f(1)}&\scriptstyle {>f(-3),}\\\scriptstyle {f(-1)}&\scriptstyle {=-6f(6).}\end{aligned}}&{\begin{aligned}\scriptstyle {f(y)}&\scriptstyle {=y^{3}-10y^{2}+31y-30,}\\\scriptstyle {f(a)}&\scriptstyle {=a^{3}-10a^{2}+31a-30,}\\\scriptstyle {f(yz)}&\scriptstyle {=y^{3}z^{3}-10y^{2}z^{2}+31yz-30,}\\\scriptstyle {f(x-2)}&\scriptstyle {=x^{3}-16x^{2}+83x-140,}\\\scriptstyle {}\end{aligned}}\end{array}}$

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

$\scriptstyle {f(2)=\phi (-2),\ f(3)=\phi (-3),\ f(5)=\phi (-4),\ f(0)+\phi (0)+246=0}$.

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

$\scriptstyle {F(0)=F(1)=F(-6)=F({\frac {1}{2}})=F(-{\frac {5}{4}})=0}$.

1. See § 14, p. 12.