Problem 47
Question
For what value(s) of \(a\) will the inequality \(a x^{2} \leq 0\) have all real numbers as its solution? Explain.
Step-by-Step Solution
Verified Answer
The inequality \(a x^{2} \leq 0\) will have all real numbers as its solution for \(a \leq 0\).
1Step 1: Understanding the problem
We are given that the inequality \(a x^{2} \leq 0\) has all real numbers as solutions. This means that the parabola \(y = a x^{2}\) is either on or below the x-axis for all real \(x\).
2Step 2: Observing the graph of \(y=x^{2}\)
The graph of \(y=x^{2}\) is a parabola that opens upwards and touches the x-axis at \(x=0\), meaning, \(x^{2} \geq 0\) for all real values of \(x\).
3Step 3: Making the inequality true
To satisfy \(a x^{2} \leq 0\), \(a\) must be non-positive. Because if \(a\) was positive, it will make the product \(a x^{2}\) positive for all non-zero \(x\), and since \(a x^{2} > 0\) is not consistent with our inequality, \(a\) must be non-positive for the inequality to hold.
4Step 4: Deriving the range of \(a\)
From the above, we know the range for which \(a\) makes the inequality \(a x^{2} \leq 0\) true for all real \(x\) is \(a \leq 0\).
Key Concepts
Real Number SolutionsParabola GraphInequality Solutions
Real Number Solutions
Real number solutions for quadratic inequalities occur when every real number satisfies the inequality. In our given inequality, \( a x^{2} \leq 0 \), we need this condition to be true for all possible values of \( x \). If \( a \) is zero, the inequality becomes \( 0 \leq 0 \), which is true for every \( x \). However, when \( a \) is negative, \( ax^2 \) results in non-positive values for all real numbers since \(x^2\) is non-negative. Thus, a key takeaway is:
- For the inequality to hold for all real numbers, \(a\) can be zero or negative.
Parabola Graph
In quadratic equations and inequalities, parabola graphs play a crucial role. A standard parabola graph is described by the function \( y = x^2 \), which is always positive or zero. It opens upward and touches the x-axis at one point: the vertex, located at \(x = 0\). For the inequality \( ax^2 \leq 0\), the parabola defined by \( y = ax^2 \) determines the solutions:
- When \(a\) is positive, the parabola opens upward, keeping above the x-axis except at \(x=0\).
- When \(a\) is zero, the graph becomes a flat line on the x-axis, satisfying the inequality for all \(x\).
- If \(a\) is negative, the parabola flips and opens downward, ensuring \( y = ax^2 \) stays at or below the x-axis.
Inequality Solutions
Finding solutions for a quadratic inequality such as \( ax^2 \leq 0 \) involves checking when the expression is at or below zero. The sign of \( a \) is critical in this analysis. If \( a \) is positive, the inequality cannot hold as \( ax^2 \), a product of two positive values, will be positive for any \( x eq 0 \). However, for inequality solutions:
- If \(a = 0\), the inequality simplifies to \(0 \leq 0\), holding for any \(x\).
- If \(a \lt 0\), \( ax^2 \) becomes non-positive, meaning it's either zero or negative across all real numbers, satisfying the inequality condition.
Other exercises in this chapter
Problem 46
Find the average rate of change of each ficnetion on the given interval. $$f(x)=x^{3}+1 ; \text { interval: }[0,2]$$
View solution Problem 47
Find the vertex and axis of symmetry of the associated parabola for each quadratic function.Then find at least two additional points on the parabola and sketch
View solution Problem 47
Solve the radical equation to find all real solutions. Check your solutions. $$\sqrt{2 x+3}-\sqrt{x-2}=2$$
View solution Problem 47
In Exercises \(41-48,\) use \(f\) and \(g\) given by the following tables of values. $$\begin{array}{ccccc}x & -1 & 0 & 3 & 6 \\\\\hline f(x) & -2 & 3 & 4 & 2\e
View solution