Use the given derivative to find all critical points of 'f' and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume that 'f' is continuous everywhere.
f' (x) = (1-2x)/ ∛(x+3)
∛(x+3)in words is the cubeth root of (x+3) if its not clear.
To find the critical points of a function, we need to set the derivative equal to zero and solve for x. In this case, the derivative of f(x) is given by:
f'(x) = (1 - 2x) / ∛(x + 3)
Setting this equal to zero:
(1 - 2x) / ∛(x + 3) = 0
To simplify the equation, we can multiply through by ∛(x + 3):
1 - 2x = 0
Now, solve for x:
2x = 1
x = 1/2
So, the critical point of f(x) occurs at x = 1/2.
To determine whether this critical point corresponds to a relative maximum, relative minimum, or neither, we can use the second derivative test. The second derivative of f(x) can be found by differentiating f'(x), which is:
f''(x) = (d/dx)((1 - 2x) / ∛(x + 3))
Before calculating the second derivative, let's simplify the expression by expanding the denominator:
f''(x) = (d/dx)((1 - 2x)(x + 3)^(-1/3))
= (d/dx)(1 - 2x)(x + 3)^(-1/3)
Using the product rule, we have:
f''(x) = (1 - 2x)(-1/3)(x + 3)^(-4/3) + (-2)(x + 3)^(-1/3)
Now, we can substitute x = 1/2 into the second derivative to evaluate it at the critical point:
f''(1/2) = (1 - 2(1/2))(-1/3)((1/2) + 3)^(-4/3) + (-2)((1/2) + 3)^(-1/3)
Simplifying further:
f''(1/2) = 0 + (-2)(4 + 3)^(-1/3)
= -2(7)^(1/3)
Since the second derivative at x = 1/2 is negative (-2(7)^(1/3) < 0), the second derivative test tells us that a relative maximum occurs at this critical point.
Therefore, the function f(x) has a relative maximum at x = 1/2.