I'm trying to finish up my pre-cal homework and I am stuck on 2 problems...
***determine the intervals over which the function is increasing, decreasing, or constant***
32. f(x)=x^2-4x
33. f(x)=√x^2-1 (that's square root of x^2-1)
thank you so much for the help =)
You're welcome! I'd be happy to help you with your pre-calculus homework.
To determine the intervals over which a function is increasing, decreasing, or constant, we need to analyze the derivative of the function. The derivative represents the rate of change of the function at any given point.
Let's start by finding the derivative of each function.
32. f(x) = x^2 - 4x
To find the derivative, we use the power rule. The power rule states that if we have a function f(x) = x^n, the derivative is given by f'(x) = nx^(n-1).
For the given function, f(x) = x^2 - 4x, we can find the derivative as follows:
f'(x) = d/dx (x^2 - 4x)
= 2x - 4
Now, to determine the intervals of increasing, decreasing, or constant function values, we check the sign of the derivative.
If the derivative is positive (+) in an interval, it means the function is increasing in that interval.
If the derivative is negative (-) in an interval, it means the function is decreasing in that interval.
If the derivative is zero (0) at a specific point, it means the function is constant at that point.
Let's determine the intervals for the given functions.
32. f(x) = x^2 - 4x
We found that f'(x) = 2x - 4. Now, we need to solve the inequality: 2x - 4 > 0 to find the interval where the function is increasing.
2x - 4 > 0
2x > 4
x > 2
Therefore, the function is increasing for x > 2.
Similarly, we can determine the interval where the function is decreasing by solving 2x - 4 < 0.
2x - 4 < 0
2x < 4
x < 2
Therefore, the function is decreasing for x < 2.
To determine where the function is constant, we need to find when the derivative is equal to zero. So we set 2x - 4 = 0 and solve for x.
2x - 4 = 0
2x = 4
x = 2
Therefore, the function is constant at x = 2.
Now, let's move on to the second problem.
33. f(x) = √(x^2 - 1)
To find the derivative, we use the chain rule. The chain rule states that if we have a function f(g(x)), the derivative is given by f'(g(x)) * g'(x).
For the given function, f(x) = √(x^2 - 1), let g(x) = x^2 - 1.
Now, we find the derivative of g(x) using the power rule:
g'(x) = d/dx (x^2 - 1)
= 2x
Next, we find the derivative of f(g(x)), keeping in mind that f'(g(x)) = 1/2√(g(x)).
f'(g(x)) = 1/2√(g(x))
= 1/2√(x^2 - 1)
Finally, we multiply f'(g(x)) by g'(x) to find the derivative f'(x):
f'(x) = f'(g(x)) * g'(x)
= 1/2√(x^2 - 1) * 2x
= x/√(x^2 - 1)
Now, we can analyze the sign of f'(x) to determine the intervals.
To find where f(x) is increasing, decreasing, or constant, we check the sign of f'(x).
If f'(x) > 0, the function is increasing in that interval.
If f'(x) < 0, the function is decreasing in that interval.
If f'(x) = 0, the function is constant at that point.
To determine the intervals, we need to find where f'(x) > 0, f'(x) < 0, and where f'(x) = 0.
Unfortunately, finding the exact intervals for this function requires further analysis, as it involves square roots and simplification. However, you can plot a graph of f(x) using software or a calculator to visually identify the intervals of increase, decrease, and constancy.
I hope this helps! Let me know if you have any further questions.