Find Intervals Where F(x) = 3x^5 - 5x^3 + 2 Increases

by Alex Johnson 54 views

Let's dive into finding the intervals where the function f(x) = 3x^5 - 5x^3 + 2 is increasing. This is a classic calculus problem that involves understanding derivatives and their relationship to the increasing or decreasing nature of a function. To tackle this, we'll need to compute the first derivative of f(x), analyze its critical points, and then determine the intervals where the derivative is positive. So, grab your calculus tools, and let’s get started!

1. Compute the First Derivative

To determine where a function is increasing, we first need to find its derivative. The derivative, denoted as f'(x), tells us the rate of change of the function at any given point. If f'(x) > 0, the function is increasing; if f'(x) < 0, the function is decreasing; and if f'(x) = 0, we have a critical point, which could be a local maximum, a local minimum, or a saddle point.

Given the function f(x) = 3x^5 - 5x^3 + 2, we'll use the power rule to find the derivative. The power rule states that if f(x) = ax^n, then f'(x) = nax^(n-1). Applying this rule to each term in our function, we get:

  • The derivative of 3x^5 is 15x^4.
  • The derivative of -5x^3 is -15x^2.
  • The derivative of 2 (a constant) is 0.

So, the first derivative f'(x) is:

f'(x) = 15x^4 - 15x^2

This derivative is crucial because it holds the key to unlocking the intervals where our function is increasing. Now that we have f'(x), the next step is to find the critical points.

2. Find Critical Points

Critical points are the points where the derivative f'(x) is either equal to zero or undefined. These points are significant because they mark potential turning points of the function – where it changes from increasing to decreasing or vice versa. In other words, critical points are the x-values where the slope of the tangent line to the function's graph is either horizontal (f'(x) = 0) or undefined.

To find the critical points, we set f'(x) = 0 and solve for x:

15x^4 - 15x^2 = 0

We can factor out 15x^2 from the equation:

15x2(x2 - 1) = 0

This gives us three factors to consider:

  1. 15x^2 = 0, which implies x = 0.
  2. (x^2 - 1) = 0, which can be further factored as (x - 1)(x + 1) = 0, giving us x = 1 and x = -1.

Thus, our critical points are x = -1, x = 0, and x = 1. These points will divide the number line into intervals, which we will analyze in the next step to determine where f(x) is increasing.

3. Determine Intervals of Increase

Now that we have the critical points, we can determine the intervals where the function f(x) is increasing. The critical points x = -1, x = 0, and x = 1 divide the real number line into four intervals: (-∞, -1), (-1, 0), (0, 1), and (1, ∞). We'll test a value within each interval in the first derivative f'(x) = 15x^4 - 15x^2 to see if f'(x) is positive or negative.

  1. Interval (-∞, -1): Choose a test value, say x = -2. f'(-2) = 15(-2)^4 - 15(-2)^2 = 15(16) - 15(4) = 240 - 60 = 180 Since f'(-2) > 0, the function is increasing on the interval (-∞, -1).

  2. Interval (-1, 0): Choose a test value, say x = -0.5. f'(-0.5) = 15(-0.5)^4 - 15(-0.5)^2 = 15(0.0625) - 15(0.25) = 0.9375 - 3.75 = -2.8125 Since f'(-0.5) < 0, the function is decreasing on the interval (-1, 0).

  3. Interval (0, 1): Choose a test value, say x = 0.5. f'(0.5) = 15(0.5)^4 - 15(0.5)^2 = 15(0.0625) - 15(0.25) = 0.9375 - 3.75 = -2.8125 Since f'(0.5) < 0, the function is decreasing on the interval (0, 1).

  4. Interval (1, ∞): Choose a test value, say x = 2. f'(2) = 15(2)^4 - 15(2)^2 = 15(16) - 15(4) = 240 - 60 = 180 Since f'(2) > 0, the function is increasing on the interval (1, ∞).

Therefore, the function f(x) = 3x^5 - 5x^3 + 2 is increasing on the intervals (-∞, -1) and (1, ∞). We include the endpoints because the function is non-decreasing at these points.

Conclusion

In summary, by computing the first derivative, finding critical points, and testing intervals, we've determined that the function f(x) = 3x^5 - 5x^3 + 2 is increasing on the intervals (-∞, -1] and [1, ∞). This process showcases the power of calculus in analyzing the behavior of functions. Understanding where a function is increasing or decreasing is a fundamental concept in calculus with wide-ranging applications in optimization problems, curve sketching, and more.

For further exploration of calculus concepts and techniques, you might find valuable resources at Khan Academy's Calculus section.