Calculus by Spivak, Chapter 1, Problem 4

Problem

Find all numbers $x$ for which

$4 -x < 3-2x$
$5-x^2<8$
$5-x^2<-2$
$(x-1)(x-3) >0$ (When is a product of two numbers positive?)
$x^2-2x+2>0$
$x^2+x+1>2$
$x^2-x+10 >16$
$x^2+x+1>0$
$(x- \pi) (x+5)(x-3) >0$
$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$
$\sqrt{x}{2} < 8$
$x + 3^x < 4$
$\frac{1}{x} + \frac{1}{1-x} > 0$
$\frac{x-1}{x+1} > 0$

Solution

$4 -x < 3-3x$

Collect the terms.

$x < -1$

$5-x^2<8$

Move terms around and combine.

$-3 < x^2$

$x^2$ will always be positive so any $\mathbb{R}$ will work.

$5-x^2<-2$

Move terms around and combine.

$7 < x^2$

If this were $x^2 = 7$ , we see that $x = \pm \sqrt{7}$ .

We looking for when $x^2$ is greater than 7, so the solution is $x < -\sqrt{7}$ or $x > \sqrt{7}$ .

$(x-1)(x-3) >0$

Our critical points are $x= 1$ and $x = 3$ .

If $x = 0$ , then we would have $- * - = + > 0$ .

If $x = 2$ , then we would have $+ * - = - < 0$ .

If $x = 4$ , then we would have $+ * + = + > 0$ .

So the inequality works when $x < 1$ or $x > 3$ .

$x^2-2x+2>0$

If this were $x^2 - 2x + 2 = 0$ , the solution would be $\frac{2 \pm \sqrt{-4}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$x^2+x+1>2$

$x^2 + x -1 >0$

If this were an equality and we solve for x, we’d get $x = \frac{-1 \pm \sqrt{5}}{2}$ . This is a real solution so the parabola does cross the x-axis and there are two solutions. The parabola faces up so we want the two extreme branches. Therefore $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$ .

$x^2-x+10 >16$

$x^2 -x-6 > 0$

$(x-3)(x+2) > 0$

The critical points are $x = 3$ and $x = -2$ .

If $x = -3$ , we would have $- * - = + > 0$ .

If $x = 0$ , we would have $- * + = - < 0$ .

If $x = 4$ , we would have $+ * + = + > 0$ .

Therefore, we want $x < -2$ or $x > 3$ .

$x^2+x+1>0$

If this were $x^2 -x + 1 = 0$ , the solution would be $\frac{-2 \pm \sqrt{-3}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$(x- \pi) (x+5)(x-3) >0$

The critical points are $x = \pi$ , $x = -5$ , and $x = 3$ .

If $x = -6$ , we would have $- * - * - = - < 0$ .

If $x = 0$ , we would have $- * + * - = + > 0$ .

If $x = 3.1$ , we would have $- * + * + = - < 0$ .

If $x = 4$ , we would have $+ * + * + = + > 0$ .

Therefore, the solution is $-5 < x < 3$ or $x > \pi$ .

$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$

The critical points are $x = \sqrt[3]{2}$ and $x = \sqrt{2}$ .

If $x = 0$ , we would have $- * - = + > 0$ .

If $x = 1.3$ , we would have $+ * - = - < 0$ .

If $x = 2$ , we would have $+ * + = + > 0$ .

The solution is $< \sqrt[3]{2}$ or $x > \sqrt{2}$ .

$\sqrt{x}{2} < 8$

2 is positive. $2^3 = 8$ . Any exponent greater than 3 should be greater than 8. So $x > 3$ .

$x + 3^x < 4$

By visual inspection, it’s clear that if $x = 1$ , $1 + 3^1 = 4$.

So any x less than 1 should be less than 4.

$x<3$

$\frac{1}{x} + \frac{1}{1-x} > 0$

The critical points are $x = 0$ and $x = 1$ .

If x is negative, then $\frac{1}{x} < 0$ and $\frac{1}{1-x} > 0$ , but the absolute value of $\frac{1}{x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
-1	-1	$\frac{1}{2}$
-2	$-\frac{1}{2}$	$\frac{1}{3}$
-3	$-\frac{1}{3}$	$\frac{1}{4}$
-4	$-\frac{1}{4}$	$\frac{1}{5}$

Therefore, when $x < 0$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

If $x > 1$ , then $\frac{1}{x} > 0$ and $\frac{1}{1-x} < 0$ , but the absolute value of $\frac{1}{1-x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
2	$\frac{1}{2}$	-1
3	$-\frac{1}{3}$	$-\frac{1}{2}$
4	$-\frac{1}{4}$	$-\frac{1}{3}$
5	$-\frac{1}{5}$	$-\frac{1}{4}$

Therefore, when $x > 1$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

So let’s consider $0 < x < 1$ . Consider $x = \frac{1}{2}$ .

$\frac{1}{\frac{1}{2}} + \frac{1}{(1 - \frac{1}{2})}$

Notice that for $0 < x < 1$ both terms must be positive. Therefore, the only solution is $0 < x < 1$ .

Here I have a disagreement which the answer key. It claims that the solution is $x > 1$ or $0 < x < 1$ . However, I don’t see how $x> 1$ could be a solution. Below is a plot of $\frac{1}{x} + \frac{1}{1-x}$ . It is clearly negative where $x > 1$ .