Question

) Consider the funсtion f:R\{−1}→R defined by f(x)=2x+1 /x+1.
i) Check that f(x) is well defined and 1−1.
2)Check that f(x) is not =2 for any x∈R.
3)Check that g:R\{2}→R given by g(x)=x−1/2-x is well defined and 1-1 Further,check that g(x)=−1for any x∈R.
(4) Check that (f◦g)(x)=x for x∈R\{2} and (g◦f)(x)=x for x∈R\{−1}.
b)Find the direction cosines of the perpendicular from the origin to the plane r·(6i+4j+2√ 3k)+2=0.

Answer 1

Answer on Question #52676 – Math – Linear Algebra

Question

a. Consider the function $f: \mathbb{R} \backslash \{-1\} \to \mathbb{R}$ defined by $f(x) = 2x + 1 / x + 1$.

1) Check that $f(x)$ is well defined and 1–1.

2) Check that $f(x)$ is not equal to 2 for any $x \in \mathbb{R}$.

3) Check that $g: \mathbb{R} \backslash \{2\} \to \mathbb{R}$ given by $g(x) = x - 1/2 - x$ is well defined and 1-1. Further, check that $g(x)$ is not equal to $-1$ for any $x \in \mathbb{R}$.

4) Check that $(f(g)(x) = x$ for $x \in \mathbb{R} \backslash \{2\}$ and $(g(f)(x) = x$ for $x \in \mathbb{R} \backslash \{-1\}$.

b. Find the direction cosines of the perpendicular from the origin to the plane $r \cdot (6i + 4j + 2\sqrt{3}k) + 2 = 0$.

Solution

a.

1) The function $f(x) = \frac{2x + 1}{x + 1}$ is defined everywhere except point $x = -1$.

Suppose that $f(x_1) = f(x_2)$, that is, $\frac{2x_1 + 1}{x_1 + 1} = \frac{2x_2 + 1}{x_2 + 1}$, which is equivalent to

$$\frac{2x_1 + 2 - 1}{x_1 + 1} = \frac{2x_2 + 2 - 1}{x_2 + 1}$$ $$2 - \frac{1}{x_1 + 1} = 2 - \frac{1}{x_2 + 1}$$

Subtract 2 from both sides of equality

$$- \frac{1}{x_1 + 1} = - \frac{1}{x_2 + 1}$$

The left-hand and right-hand sides of equality are equal, numerators in both sides are equal, hence denominators in both sides should be equal, that is,

$$x_1 + 1 = x_2 + 1$$

Subtract 1 from both sides and obtain

$$x_1 = x_2.$$

If $f(x_1) = f(x_2)$ implies that $x_1 = x_2$, then $f$ is one-to-one function. This statement was proved before.

Finally, $f(x) = \frac{2x + 1}{x + 1}$ is one-to-one function.

2)

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

So, one can easily see that $f(x)$ is not equal to 2 for any $x \in \mathbb{R}$, because $\frac{1}{x + 1}$ is never equal to 0.

3)

Function $g(x) = \frac{x - 1}{2 - x}$ is defined everywhere except point $x = 2$.

Suppose that $g(x_{1}) = g(x_{2})$, that is, $\frac{x_1 - 1}{2 - x_1} = \frac{x_2 - 1}{2 - x_2}$, which is equivalent to

$$\begin{array}{l} - \frac{x_1 - 1}{x_1 - 2} = - \frac{x_2 - 1}{x_2 - 2} \\ - \frac{x_1 - 2 + 1}{x_1 - 2} = - \frac{x_2 - 2 + 1}{x_2 - 2} \\ - 1 - \frac{1}{x_1 - 2} = - 1 - \frac{1}{x_2 - 2} \\ \end{array}$$

Add 1 to both sides of equality

$$- \frac{1}{x_1 - 2} = - \frac{1}{x_2 - 2}$$

The left-hand and right-hand sides of equality are equal, numerators in both sides are equal, hence denominators in both sides should be equal, that is,

$$x_1 - 2 = x_2 - 2$$

Add 2 to both sides and obtain

$$x_1 = x_2.$$

If $g(x_{1}) = g(x_{2})$ implies that $x_{1} = x_{2}$, then $g$ is one-to-one function. This statement was proved before.

Finally, $g(x) = \frac{x - 1}{2 - x}$ is one-to-one function.

4)

$$\begin{array}{l} (fog)(x) = f(g(x)) = \frac{2g(x) + 1}{g(x) + 1} = \frac{2 \cdot \frac{x - 1}{2 - x} + 1}{\frac{x - 1}{2 - x} + 1} = \frac{\frac{2 \cdot (x - 1) + 1 \cdot (2 - x)}{2 - x}}{\frac{x - 1 + 1 \cdot (2 - x)}{2 - x}} \\ = \frac{\frac{2x - 2 + 2 - x}{2 - x}}{\frac{x - 1 + 2 - x}{2 - x}} = \frac{\frac{x}{2 - x}}{\frac{1}{2 - x}} = x. \end{array}$$ $$\begin{array}{l} (gof)(x) = g(f(x)) = \frac{f(x) - 1}{2 - f(x)} = \frac{\frac{2x + 1}{x + 1} - 1}{2 - \frac{2x + 1}{x + 1}} = \frac{\frac{2x + 1 - (x + 1)}{x + 1}}{\frac{2(x + 1) - (2x + 1)}{x + 1}} = \frac{\frac{2x + 1 - x - 1}{x + 1}}{\frac{2x + 2 - 2x - 1}{x + 1}} \\ = \frac{\frac{x}{x + 1}}{\frac{1}{x + 1}} = x. \end{array}$$

b) Normal vector for this plane $r \cdot (6i + 4j + 2\sqrt{3} k) + 2 = 0$ is $(6, 4, 2\sqrt{3})$, its length is

$$\left| (6, 4, 2\sqrt{3}) \right| = \sqrt{36 + 16 + 12} = 8.$$

The direction cosines of the perpendicular from the origin to the plane $r \cdot (6i + 4j + 2\sqrt{3} k) + 2 = 0$ are

$$\frac{6}{8} = \frac{3}{4}, \frac{4}{8} = \frac{1}{2} \text{ and } \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4}.$$

Answer: a) Yes; b) $\left(\frac{3}{4}, \frac{1}{2}, \frac{\sqrt{3}}{4}\right)$.

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