1.1 Answer

1.1

(a) For example, conside the case $x = 4.0$ . Then the $y$ satisfying $y^{2} = 4.0$ are given by

$\displaystyle y^{2} - 4.0 = (y + 2.0)(y - 2.0) = 0$

and

．Then，

$\displaystyle y^{2} - x = (y + \sqrt{x})(y - \sqrt{x}) = 0$

Thus, the

satisfying $y^{2} = x, \ x > 0$ are $y = +\sqrt{x}$ and $y = -\sqrt{x}$ ．Therefore，for different $x$

，there are 2 $y$

corresopobds to it. Thus, double valued function.

(b) For example，consider $y^{3} = 2^{2}$ . Then

$\displaystyle y^{3} - 4 = y^{3} - (2^{2/3})^{3} = (y - 2^{2/3})(y^{2} + 2^{2/3}y + 2^{4/3})$

Here $y^{2} + 2^{2/3}y + 2^{4/3} = 0$ is a quadratic equation in $y$

. So, we use the discriminant. Then

$\displaystyle D = b^{2} - 4ac = (2^{2/3})^{2} - 4(1)(2^{4/3}) = 2^{4/3} - (4)2^{4/3} = 2^{4/3}(1 - 4) < 0$

Thus, this equation has no real root. Now

$\displaystyle y^{3} - x^{2} = (y - x^{2/3})(y^{2} + x^{2/3}y + x^{4/3}) = 0$

Here, $y^{2} + x^{2/3}y + x^{4/3} = 0$ has no real root. Then $\displaystyle{y = x^{\frac{2}{3}}}$ is the only real root．Therefore, single valued function.

(a) The domain of f(x) is the set of real numbers of $x$ such that $f(x)$ is real. Then $D(f) = [-2,2]$ .

(b) The domain of f(x) is the set of real numbers of $x$ such that $f(x)$ is real. Then $\displaystyle{D(h) = (-\infty,-\frac{1}{2}) \cup (0, \frac{4}{3}]}$

(a)

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

$\displaystyle (g \circ f)(x) = g(f(x)) = g(2x - 1) = (2x-1)^2 + 1 = 4x^2 - 4x + 2$

(b) $f \circ g$ cna be defined if the range of $g(x)$ is in the domain of $f(x)$ ．So, we check the range of $g(x)$ ．

$\displaystyle g(x) = \left\{\begin{array}{cl} -x, & x < 1\\ 1 + x, & x \geq 1 \end{array} \right.$

Then for

，

．Thus the the range of $g(x)$

is contained in the domain of $f(x)$

. Now we divide the case $g(x) > 0$

and $g(x) \leq 0$ .

For $x < 0$ ， $g(x) > 0$ and for $0 \leq x < 1$ ， $g(x) \leq 0$ ．Thus, for $x < 0$ ，we use $f(x) = x^2$ and for $0 \leq x < 1$ ，ew use $f(x) = 1 - x$ . Therefore,

$\displaystyle f(g(x)) = g(x)^2 = (-x)^2 = x^2, \ x < 0$

$\displaystyle f(g(x)) = 1 - g(x) = 1 - (-x) = 1 + x, \ 0 \leq x < 1$

Next for $x \geq 1$ ， $g(x) = 1 + x > 2$

．This is in the domain of $f(x)$

. Thus we use $f(x) = 1 - x, \ x \leq 0$ ．Therefore,

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

Putting together,

$\displaystyle (f \circ g)(x) = \left\{\begin{array}{cl} x^2 & x < 0\\ 1 + x & 0 \leq x < 1\\ (1 + x)^2 & x \geq 1 \end{array} \right.$

Similarly, ，

$\displaystyle (g \circ f)(x) = \left\{\begin{array}{cl} 2 - x & x \leq 0\\ - x^2 & 0 < x < 1\\ 1 + x^2 & x \geq 1 \end{array} \right.$

(a) First we show that this function is oe-to-one.

$\displaystyle f(x_{1}) = f(x_{2})$	$\displaystyle \Longrightarrow$	$\displaystyle \frac{1}{x_{1} + 2} = \frac{1}{x_{2} + 2}$
	$\displaystyle \Longrightarrow$	$\displaystyle x_{1} + 2 = x_{1} + 2$
	$\displaystyle \Longrightarrow$	$\displaystyle x_{1} = x_{2}$

Next we use $f(f^{-1}(x)) = x$ to find the inverse．

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

Solve this for $y$

. Then $\displaystyle{y + 2 = \frac{1}{x}}$ ．Therefore, $\displaystyle{y = f^{-1}(x) = \frac{1}{x} - 2}$ ．

(b) This function is not one-to-one. Since $f(x) = x^2 + 4x - 2 = (x + 2)^2 - 6$ ，it is symmetric about $y$ -axis．Thus, there is no inverse function. But we can restrict the domain of $f(x)$ so that it is one-to-one. For if we write the domain as $\{x < -2\} \cup \{x \geq 2\}$ ， then on this domain, we can find the inverse.

$\displaystyle f(f^{-1}(x)) = f(y) = y^2 + 4y - 2 = x$

Solve this for $y$

. Then we have $\displaystyle{y = -2 + \sqrt{x + 6} \ (x \geq -6) \ y = -2 - \sqrt{x + 6} \ (x \geq -6)}$