if f(x)=(x-3)^4(2x+1)^3 find the value(s) of x which solve f'(x)=0
f = (x-3)^4 * (2x+1)^3
f' = 4*(x-3)^3 * (2x+1)^3 + (x-3)^4 * 3 * (2x+1)^2 * 2
= 4(x-3)^3(2x+1)^3 + 6(x-3)^4(2x+1)^2
= 2(x-3)^3(2x+1)^2 (2(x-3) + 3(x-3))
= 2(x-3)^3(2x+1)^2(2x-6+3x-9)
= 2(x-3)^3(2x+1)^2(5x-15)
= 10(x-3)^4(2x+1)^2
So, f'=0 when x = 3 or -1/2
f = (x-3)^4 * (2x+1)^3
f' = 4*(x-3)^3 * (2x+1)^3 + (x-3)^4 * 3 * (2x+1)^2 * 2
= 4(x-3)^3(2x+1)^3 + 6(x-3)^4(2x+1)^2
= 2(x-3)^3(2x+1)^2 (2(2x+1) + 3(x-3))
= 2(x-3)^3(2x+1)^2(4x+2+3x-9)
= 2(x-3)^3(2x+1)^2(7x-7)
= 14(x-3)^3(2x+1)^2(x-1)
So f' = 0 when x = -1/2 or 1 or 3
To find the value(s) of x that solve f'(x) = 0, we need to differentiate the function f(x) and set the derivative equal to zero.
First, let's find the derivative of f(x) using the product and chain rules. The derivative of f(x) with respect to x, f'(x), can be found as follows:
f(x) = (x - 3)^4 * (2x + 1)^3
Let's consider each part separately:
1. The derivative of (x - 3)^4:
Using the chain rule, we multiply the derivative of the inner function (x - 3) by the derivative of the outer function (x - 3)^3.
The derivative of (x - 3) is simply 1, so we have: 4(x - 3)^3.
2. The derivative of (2x + 1)^3:
Applying the chain rule again, we multiply the derivative of the inner function (2x + 1) by the derivative of the outer function (2x + 1)^2.
The derivative of (2x + 1) is 2, so we have: 3(2x + 1)^2 * 2.
Now, we can find the derivative of f(x) by multiplying these two parts together:
f'(x) = 4(x - 3)^3 * 3(2x + 1)^2 * 2
Now, set f'(x) equal to zero and solve for x:
0 = 4(x - 3)^3 * 3(2x + 1)^2 * 2
Since the product of several factors is equal to zero, at least one of the factors must be equal to zero.
To find the values of x that satisfy the equation f'(x) = 0, we can set each factor equal to zero separately and solve for x.
Factor 1: 4(x - 3)^3 = 0
Solve for x:
(x - 3)^3 = 0
Taking the cube root of both sides, we get:
x - 3 = 0
x = 3
Factor 2: 3(2x + 1)^2 = 0
Solve for x:
(2x + 1)^2 = 0
Taking the square root of both sides, we get:
2x + 1 = 0
2x = -1
x = -1/2
Therefore, the values of x that solve f'(x) = 0 are x = 3 and x = -1/2.