Porisma

Worked solution

A repeated zero and an exponential deduction

Without the use of a calculator, solve the inequality 1x+1≤x(1+x)\dfrac1x+1\leq x(1+x). marks

Hence, solve 1−e2x+e−x−ex≤01-e^{2x}+e^{-x}-e^x\leq0. marks

[2013/DHS/I/2]

CJC · Review Problems · Question 4

Answer

Base: (−∞,0)∪[1,∞)(-\infty,0)\cup[1,\infty). Hence x≥0x\geq0.

First inequality

Move the right side to the left and factor: 1x+1−x(1+x)=1+x−x2−x3x=(1+x)2(1−x)x≤0\dfrac1x+1-x(1+x)=\dfrac{1+x-x^2-x^3}{x}=\dfrac{(1+x)^2(1-x)}x\leq0. The signs across −1,0,1-1,0,1 are −,−,+,−-,-,+,-. The repeated numerator zero −1-1 is included without a sign change. The pole 0 is excluded and the zero 1 is included. Thus x<0x<0 or x≥1x\geq1.

How did this part go?

Hence

Let the original input be t=ex>0t=e^x>0. Its left-minus-right expression is 1t+1−t(1+t)=1−e2x+e−x−ex\dfrac1t+1-t(1+t)=1-e^{2x}+e^{-x}-e^x. The original result requires t<0t<0 or t≥1t\geq1. Reject t<0t<0 because ex>0e^x>0; ex≥1e^x\geq1 gives x≥0x\geq0.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

The expression (1+x)²(1−x)/x is negative on both sides of −1 until the pole 0, positive on (0,1), then negative. Include −1 and 1; exclude 0.
The expression (1+x)²(1−x)/x is negative on both sides of −1 until the pole 0, positive on (0,1), then negative. Include −1 and 1; exclude 0.
The expression (1+x)²(1−x)/x is negative on both sides of −1 until the pole 0, positive on (0,1), then negative. Include −1 and 1; exclude 0.

Common mistakes

A repeated root can be included even when the sign does not change. The substituted exponential input cannot be negative.

Solution notes

The sign chart is independently drawn from the printed solution. Spacing is illustrative; labelled critical values and open/closed endpoints determine the set.

Read the hint