Problem 36

Question

In Exercises $34-39,$ solve the inequality analytically. $$ 2^{\left(x^{3}-x\right)}<1 $$

Step-by-Step Solution

Verified

Answer

The solution set is $(-\infty, -1) \cup (0, 1)$.

1Step 1: Understand the Inequality

We need to solve the inequality $2^{(x^3 - x)} < 1$. Since $2^0 = 1$, we understand that the exponent $x^3 - x$ must be less than zero for the inequality to hold. This leads us to the inequality $x^3 - x < 0$.

2Step 2: Factor the Cubic Expression

Consider the expression $x^3 - x$. We can factor it by taking $x$ out as a common factor: $x(x^2 - 1) < 0$. Further, factor $x^2 - 1$ as the difference of squares: $x(x - 1)(x + 1) < 0$.

3Step 3: Identify Critical Points

The critical points of the inequality $x(x-1)(x+1) = 0$ are where the expression is zero. Solving $x(x-1)(x+1) = 0$ gives us $x = 0$, $x = 1$, and $x = -1$. These are the points where the sign of the expression can change.

4Step 4: Test Intervals Between Critical Points

We need to test the sign of $x(x-1)(x+1)$ in the intervals determined by the critical points $x = -1$, $x = 0$, and $x = 1$:- For $x < -1$, choose $x = -2$: $(-2)((-2)-1)((-2)+1) = (-2)(-3)(-1) = -6$, negative.- For $-1 < x < 0$, choose $x = -0.5$: $(-0.5)((-0.5)-1)((-0.5)+1) = (-0.5)(-1.5)(0.5) = 0.375$, positive.- For $0 < x < 1$, choose $x = 0.5$: $(0.5)((0.5)-1)((0.5)+1) = (0.5)(-0.5)(1.5) = -0.375$, negative.- For $x > 1$, choose $x = 2$: $(2)(2-1)(2+1) = (2)(1)(3) = 6$, positive.

5Step 5: Determine the Solution Set

Based on the sign of the expression in each interval, the inequality $x(x-1)(x+1) < 0$ holds for the intervals where the sign is negative. From our testing, the negative intervals are $(-\infty, -1)$ and $(0, 1)$.

Key Concepts

FactorizationCubic ExpressionsCritical PointsTest Intervals

Factorization

Factorization is a mathematical process used to express an expression as a product of its factors. In our inequality problem, we start by examining the expression $x^3 - x$.
By factoring, we aim to simplify it to find its roots easily, which helps in solving the inequality.

Firstly, notice that $x$ is a common factor in each term of $x^3 - x$. We factor $x$ out to get $x(x^2 - 1)$.
The next step is to factor $x^2 - 1$. This is a difference of squares, which can be rewritten as $(x - 1)(x + 1)$.
Thus, the original expression $x^3 - x$ factors into $x(x - 1)(x + 1)$.

This factorized form is crucial as it gives us potential points where the sign of the inequality might change.

Cubic Expressions

Cubic expressions are polynomials of degree three, generally in the form $ax^3 + bx^2 + cx + d$. In this task, we have $x^3 - x$, which can influence how an inequality behaves as it factors into simpler components.
Cubic expressions can have:

Up to three real roots.
Different turning points where the graph of the expression changes direction.

By analyzing the cubic expression $x^3 - x$, we simplify our task by reducing it to evaluating its roots and the intervals around them, using the factorization method, which forms the basis of our solution strategy.

Critical Points

Critical points are the values of $x$ where the value of the expression changes, usually crossing the x-axis. For the factored expression $x(x - 1)(x + 1)$, the critical points occur where each factor equals zero.

Solve each factor for zero: $x = 0$, $x - 1 = 0$, and $x + 1 = 0$, resulting in the critical points $x = 0$, $x = 1$, and $x = -1$.
These points divide the x-axis into intervals where the sign of the expression might change.

Critical points are pivotal since they indicate potential changes in the solution set of an inequality.

Test Intervals

Testing intervals involve checking the sign of a factored expression in different intervals determined by the critical points. This tells us where the inequality holds true.
After identifying critical points at $x = -1$, $x = 0$, and $x = 1$, we analyze intervals: