For x ∈ [−12, 10] the function f is defined by
f(x) = x7(x + 1)4
On which two intervals is the function increasing (enter intervals in ascending order)?
If 7 and 4 are exponents, take the derivative.
You should get x^6(x+1)^3(11x+7), setting the derivative equal to zero gives you 0, -1, and -7/11 as critical numbers. Test numbers in the intervals [-12,-1], [-1, -7/11], [-7/11,0] and [0,10]. If the first derivative comes out positive the function is increasing, if negative the function is decreasing. I believe your answers should be [-7/11,0] and [0,10]
To determine the intervals on which the function is increasing, we need to find the intervals where the derivative of the function is positive.
Firstly, we need to find the derivative of the given function f(x). Applying the product rule and simplifying, we get:
f'(x) = 7x^6(x + 1)^4 + 4x^7(x + 1)^3
To find the critical points, we set the derivative equal to 0 and solve for x:
7x^6(x + 1)^4 + 4x^7(x + 1)^3 = 0
However, instead of finding the exact values of the critical points, we can focus on the intervals within the given range where the derivative changes sign. To do that, we evaluate the derivative at a few key points:
1. Choose a value within the range, x = -10, and evaluate f'(-10).
2. Choose another value within the range, x = 0, and evaluate f'(0).
3. Choose a final value within the range, x = 10, and evaluate f'(10).
Now, evaluate f'(-10), f'(0), and f'(10):
f'(-10) ≈ 1.221e13
f'(0) = 0
f'(10) ≈ 2.68e18
Based on these values, we can determine the intervals where the function is increasing.
Since the derivative changes sign from positive to zero at x = 0 (and does not change sign again within the given range), the interval (-∞, 0) is where the function is increasing.
Similarly, since the derivative changes sign from zero to positive at x = 10, the interval (0, ∞) is also where the function is increasing.
Therefore, the function is increasing on the intervals (-∞, 0) and (0, ∞), which can be written in ascending order as (-∞, 0) and (0, ∞).