Suppose the derivative of a function f is f′(x)=(x−5)^3(x−3)^4(x+18)^8. Then the function f is increasing on the interval
Can anyone please help me with this problem. It seems simple, but I just do not understand it.
f' is the slope of f
if f' > 0, then the curve is sloping upward
That is, f is increasing
So, we need to find where f' > 0
Since (x−3)^4(x+18)^8 is always positive (even powers are never negative) except at x=3 or x=-18
then f' > 0 where (x-5) > 0
That is, where x > 5
Since neither of the above two values of x, where f' = 0 is in the interval x > 5, then we are done. See the graph at
http://www.wolframalpha.com/input/?i=integral+%28x%E2%88%925%29%5E3%28x%E2%88%923%29%5E4%28x%2B18%29%5E8+dx
The curve is so steep it's hard to see, but you can tell that for x>5 the curve is sloping upward.
To determine when a function is increasing or decreasing, we need to find the intervals where the derivative is positive or negative.
Given the derivative f′(x) = (x − 5)^3(x − 3)^4(x + 18)^8, we know that the function f is increasing whenever f′(x) > 0 and decreasing whenever f′(x) < 0.
Let's break down the intervals:
1. (x - 5)^3:
- This factor equals zero when x = 5.
- When x < 5, this factor is negative.
- When x > 5, this factor is positive.
2. (x - 3)^4:
- This factor equals zero when x = 3.
- When x < 3, this factor is negative.
- When x > 3, this factor is positive.
3. (x + 18)^8:
- This factor never equals zero since it is raised to an even power.
- When x < -18, this factor is positive.
- When x > -18, this factor is positive.
Now, we can combine the information from all of the factors:
- When x < -18, all three factors are negative, so f′(x) < 0.
- When -18 < x < 3, (x - 5)^3 and (x - 3)^4 are positive, but (x + 18)^8 is still positive, so f′(x) > 0.
- When 3 < x < 5, (x - 5)^3 is negative, but (x - 3)^4 and (x + 18)^8 are positive, so f′(x) < 0.
- When x > 5, all three factors are positive, so f′(x) > 0.
Therefore, the function f is increasing on the interval (-∞, -18) and (3, ∞).