Final answer:
To find the critical points of the function x⁴(x-3)³, we need to find the values of x where the derivative of the function is equal to zero or undefined.
Step-by-step explanation:
To find critical points of a given function, we need to follow these steps:
1. Take the derivative of the function.
2. Set the derivative equal to zero and solve for x to find potential critical points.
3. Test if the function is defined for each value of x obtained in step 2 to confirm if they are indeed critical points. Let's find the critical points for the function: f(x) = x⁴(x-3)³
Step 1: Find the derivative of f(x) To differentiate this, we'll use the product rule. The product rule states that the derivative of a product of two functions is: d/dx [u(x) * v(x)] = u'(x)v(x) + u(x)v'(x) Let's differentiate the two functions separately and then apply the product rule: Let u(x) = x⁴ then u'(x) = 4x³ Let v(x) = (x-3)³ then v'(x) can be found by applying the chain rule: For (x-3)³ we have an outer function g(t) = t³ and inner function t(x) = x - 3 g'(t) = 3t² and t'(x) = 1 So v'(x) = g'(t(x)) * t'(x) = 3(x-3)² * 1 = 3(x-3)²
Now, apply the product rule: f'(x) = u'(x)v(x) + u(x)v'(x) = 4x³(x-3)³ + x⁴*3(x-3)²
Step 2: Set the derivative equal to zero Now we want to find the values of x where f'(x) = 0: 0 = 4x³(x-3)³ + x⁴*3(x-3)²
We can make this equation easier to solve by factoring out the common terms: 0 = x³(x-3)²[4(x-3) + 3x] Solve for x in each factor separately: From x³ = 0 : we get x = 0 From (x-3)² = 0 : we get x = 3 (since squaring any real number cannot give a negative result, so the only solution here is x = 3) From 4(x-3) + 3x = 0 : 4x - 12 + 3x = 0 7x - 12 = 0 7x = 12 x = 12/7 (approximately 1.714)
Now we have our critical points: x = 0, x = 3, and x = 12/7.
Step 3: Confirm critical points Last step involves checking if the function is defined for all our critical points.
The function f(x) is a polynomial and is defined everywhere on the real number line, so all values we just found are indeed valid critical points. Thus, the critical points of the function f(x) = x⁴(x-3)³ are x = 0, x = 3, and x = 12/7.