5x / x^2+1 less then or equal to x. Please someone help help been stuck on it for a while and I have an exam tommorrow. Any help is greatly appreciated
If you mean: (5x) / (x^2 + 1),
Then it depends entirely on what value of 'x' you take.
For x = 0, the two terms are equal.
For x = 1, (5x) / (x^2 + 1) is greater than one.
For x = -1, (5x) / (x^2 + 1) is less than one.
5x/(x^2+1) <= x.
Multiply both sides by x^2 + 1:
5x<= x(x^2+1),
5 <= x^2 + 1,
4 <= x^2
2 <= X.
X=> 2.
To solve the inequality \( \frac{5x}{{x^2+1}} \leq x \), we can follow these steps:
Step 1: Multiply both sides of the inequality by \( (x^2+1) \) to eliminate the denominator:
\( 5x \leq x(x^2+1) \)
Step 2: Distribute on the right side:
\( 5x \leq x^3 + x \)
Step 3: Rearrange the terms to have the inequality on one side and zero on the other:
\( x^3 - 4x \geq 0 \)
Now, to find the critical points where the inequality may change direction, we need to find the values of x that make the left side equal to zero. In this case, we have a cubic polynomial, so we can find its critical points by factoring.
Step 4: Factor out an x to get:
\( x(x^2 - 4) \)
Step 5: Further factor the quadratic expression:
\( x(x - 2)(x + 2) \)
Now we have the critical points \( x = 0 \), \( x = 2 \), and \( x = -2 \). These points divide the number line into four intervals: \( (-\infty, -2) \), \( (-2, 0) \), \( (0, 2) \), and \( (2, \infty) \).
Step 6: Test a value from each interval to determine the sign of the inequality.
For example, in the interval \( (-\infty, -2) \), we can test \( x = -3 \):
Plugging it into the original inequality:
\( \frac{5(-3)}{(-3)^2 + 1} \leq -3 \)
\( -\frac{15}{10} \leq -3 \)
\( -\frac{3}{2} \leq -3 \)
This is true, so the inequality holds in the interval \( (-\infty, -2) \).
By performing similar tests, you will find that the inequality holds in the intervals \( (-2, 0) \) and \( (2, \infty) \) as well.
Step 7: To determine the intervals where the inequality does not hold, you can test a value within the interval \( (0, 2) \). Let's test \( x = 1 \):
Plugging it into the original inequality:
\( \frac{5(1)}{(1)^2 + 1} \leq 1 \)
\( \frac{5}{2} \leq 1 \)
This is false, so the inequality does not hold in the interval \( (0, 2) \).
In summary, the solution to the inequality \( \frac{5x}{{x^2+1}} \leq x \) is:
\( x \in (-\infty, -2] \cup (-2, 0] \cup (2, \infty) \)