Final answer:
The function f(x) = (x²)⁵(x−3)⁴ has 2 critical points. These occur at the values for which the derivative is zero, which are x = 0 (with a multiplicity of 5) and x = 3 (with a multiplicity of 4), but multiplicity does not affect the count of distinct critical points. The correct answer is 2 which is not mentioned in the options.
Step-by-step explanation:
The function given is f(x) = (x²)⁵(x−3)⁴. To find the critical points of this function, we need to find the points at which the first derivative is equal to zero or is undefined. The critical points occur when the factors of the derivative are equal to zero.
Let's find the derivative of f(x). Using the product rule and the chain rule, we have:
f'(x) = 5(x²)⁴·2x(x−3)⁴ + (x²)⁵·4(x−3)³
This simplifies to:
f'(x) = 10x⁵(x−3)⁴ + 4x²²4 (x²)⁴(x−3)
The critical points occur where f'(x) = 0. Thus, we set each factor equal to zero:
- x = 0
- x = 3
Notice that x = 0 makes the term 10x⁵ equal to zero, and x = 3 makes the term (x−3)⁴ equal to zero. Since we have raised x² to the power of 5 and (x−3) to the power of 4, the multiplicity of each factor must also be considered. The factor x² will be zero for two values of x: x = 0 and x = -0 (which is still 0). The factor (x−3) will be zero for one value of x: x = 3.
Therefore, considering the multiplicity of each factor, x = 0 has multiplicity 5 (since it's raised to the power of 5), and x = 3 has multiplicity 4 (since it's raised to the power of 4). However, even if the power is high, the zero only counts as one distinct critical point. So, we have a total of 2 critical points: one at x = 0 and one at x = 3.