for the function y=-(1)/(x-3/4) describe the intervals where the slope is increasing and where its decreasing
look for where second derivative is positive. ( x <0.75)
https://www.wolframalpha.com/input/?i=second+derivative+of+-1%2F%28x-.75%29
To determine the intervals where the slope is increasing and decreasing for the function y = -(1)/(x - 3/4), we need to find the derivative of the function and examine its sign changes.
Step 1: Find the derivative of the function y = -(1)/(x - 3/4).
To find the derivative, we will use the quotient rule.
Let's denote the function as f(x) = -(1)/(x - 3/4).
Using the quotient rule, we have:
f'(x) = [(0)(x - 3/4) - (-1)(1)]/(x - 3/4)^2
= 1/(x - 3/4)^2
Step 2: Determine where the derivative is positive or negative.
To find the intervals where the slope is increasing or decreasing, we need to determine where the derivative is positive or negative.
Since f'(x) = 1/(x - 3/4)^2, the denominator (x - 3/4)^2 is always positive. Hence, the sign of the derivative depends only on the numerator, which is 1.
When 1 is positive, f'(x) = 1/(x - 3/4)^2 is positive.
When 1 is negative, f'(x) = 1/(x - 3/4)^2 is negative.
Since 1 is always positive, f'(x) = 1/(x - 3/4)^2 is always positive.
Therefore, the slope is always increasing for the function y = -(1)/(x - 3/4).
To determine the intervals where the slope of the function y = -(1)/(x - 3/4) is increasing or decreasing, we need to analyze the sign of the derivative of the function.
First, let's find the derivative of the function with respect to x:
dy/dx = d/dx [-(1)/(x - 3/4)]
To find the derivative, we can use the quotient rule. The quotient rule states that for a function u/v, the derivative is (v * du/dx - u * dv/dx) / (v^2).
So, applying the quotient rule to our function, we have:
dy/dx = [0 * (x - 3/4) - (-1) * 1] / (x - 3/4)^2
dy/dx = 1 / (x - 3/4)^2
Now that we have the derivative, we need to analyze its sign to determine where the slope is increasing or decreasing. For this, we will solve the inequality:
dy/dx > 0
Let's find the intervals where the derivative is positive (dy/dx > 0):
1 / (x - 3/4)^2 > 0
Since the denominator (x - 3/4)^2 is always positive, the inequality becomes:
1 > 0
This inequality is true for all x values. Therefore, the slope is increasing for all x values. In other words, the function has a positive slope everywhere.
To summarize:
- The slope of the function y = -(1)/(x - 3/4) is increasing for all x values.
- The function has a positive slope everywhere.