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Suppose the derivate of a function f is f'(x)= (x+)² (x-4)³ (x-7)⁴. On what interval is f increasing?

User Berenbums
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Final answer:

The function f is increasing on the intervals (-∞, 4) and (7, ∞). The derivative f'(x)= (x+)² (x-4)³ (x-7)⁴ indicates the function is increasing everywhere except between x=4 and x=7, where it is decreasing.

Step-by-step explanation:

Intervals Where Function f is Increasing

To determine on what interval the function f is increasing, we must look at the sign of the derivative f'(x). Since f'(x) is given as f'(x) = (x+)² (x-4)³ (x-7)⁴, we need to find the critical points where the derivative is zero or undefined. The critical points of this function are x = - (because the positive sign is likely a typo, this could be meant as x = 0, but can't be determined), x = 4, and x = 7.

Since the exponent of each factor is even, these factors always yield non-negative results. Therefore, to determine the intervals of increase, we consider the sign of the derivative between critical points:


  • Between negative infinity and 0, the derivative is positive (since all exponents are even).

  • Between 0 and 4, the derivative is positive.

  • Between 4 and 7, the derivative becomes negative.

  • Between 7 and infinity, the derivative is positive.

Consequently, the function f is increasing on the intervals (-∞, 4) and (7, ∞). As the derivative is negative on the interval (4, 7), the function is decreasing there.

User Adam Biggs
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