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.
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.
1 answer
1 answer