1.1 Answer

1.1

1.

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

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

and $y = 2.0$ or $y = -2.0$．Then，

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

Thus, the $y$ 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

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

Here $y^{2} + 2^{2/3}y + 2^{4/3} = 0$ is a quadratic equation in $y$. So, we use the discriminant. Then

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

Thus, this equation has no real root. Now

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

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

2.

(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 $${D(h) = (-\infty,-\frac{1}{2}) \cup (0, \frac{4}{3}]}$$

3.

(a)

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

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

(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 $g(x)$．

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

Then for $x < 1$， $g(x) = - x > -1$．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,

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

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

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,

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

Putting together,

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

Similarly, ，

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

4.

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

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

$$\Longrightarrow x_{1} + 2 = x_{1} + 2$$

$$\Longrightarrow x_{1} = x_{2}$$

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

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

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

(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.

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

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