Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x(1/3) + 6x(4/3). You must justify your answer using an analysis of f '(x) and f "(x).
My start to a solution:
x-values: (-1/8) and (-1/4)
I got -1/8 from f'(x)- extrema
I got -1/4 from f"(x)- inflection point
I'm not sure where to go from here
I will assume those were exponents and you want:
f(x) = 3x^(1/3) + 6x^(4/3)
f ' (x) = x^(-2/3) + 8x^(1/3)
f '' (x) = (-2/3)x^(-5/3) + (8/3)x^(-2/3)
checking your work:
x^(-2/3) (1 + 8x) = 0
x = -1/8 , you had that!
but that is only the x of the extrema point
so f(-1/8) = 3(-1/8)^(1/3) + 6(-1/8)^(4/3)
= 3(-1/2) + 6(-1/16) = -15/8
so the extrema point is (-1/8, -15/8)
but right now we don't know if it is a max or a min
f ''(x) = 0 for points of inflection
(-2/3)x^(-5/3) + (8/3)x^(-2/3) = 0
x^(-5/3) - 4x^(-2/3) = 0
x^(-5/3)(1 - 4x) = 0
x = 1/4
You had -1/4
again , all you have is the x, we need the y
f(1/4) = .....
I will let you do that, not going to be "exact" since cube root of 4 is irrational
also f '' (-1/8) = positive, so (-1/8, -15/8) is a MINIMUM point.
You might want to check my arithmetic, I should have written it out on paper.
To find the x-coordinates of relative extrema and inflection points, you need to analyze the derivative of the function, f'(x), and the second derivative, f''(x).
First, let's find the derivative of f(x):
f(x) = 3x^(1/3) + 6x^(4/3)
To find f'(x), we can use the power rule: if f(x) = cx^n, then f'(x) = cnx^(n-1).
f'(x) = (3/3)x^(-2/3) + (6/3)x^(1/3)
= x^(-2/3) + 2x^(1/3)
Now, let's find the x-coordinates of the relative extrema by solving for f'(x) = 0:
x^(-2/3) + 2x^(1/3) = 0
To solve this equation, we can introduce a common denominator on the left side:
(x^(1/3))/x + 2x^(1/3) = 0
(x^(1/3) + 2x^(4/3))/x = 0
Since the numerator must equal 0, we have:
x^(1/3) + 2x^(4/3) = 0
Now, we can factor out x^(1/3):
x^(1/3)(1 + 2x) = 0
To find the values of x, we set each factor equal to 0:
x^(1/3) = 0 or 1 + 2x = 0
From the first equation, we can see that x^(1/3) = 0 has no real solution, as 0 raised to any power is still 0.
Let's solve the second equation:
1 + 2x = 0
2x = -1
x = -1/2
So, we have one relative extremum at x = -1/2.
Now, let's find the second derivative, f''(x), to determine any inflection points.
To find f''(x), we differentiate f'(x):
f''(x) = d/dx (x^(-2/3) + 2x^(1/3))
= (-2/3)x^(-5/3) + (2/3)x^(-2/3)
We want to find the x-coordinates where f''(x) = 0 or does not exist.
Setting f''(x) = 0:
(-2/3)x^(-5/3) + (2/3)x^(-2/3) = 0
Multiplying through by 3x^(5/3):
-2 + 2x^3 = 0
Rearranging the equation:
2x^3 = 2
x^3 = 1
x = 1
Therefore, we have one inflection point at x = 1.
In summary:
- The x-coordinate of the relative extremum is: x = -1/2
- The x-coordinate of the inflection point is: x = 1
Please note that this analysis assumes that the function is defined within the given domain.
To find the x-coordinates of the relative extrema and inflection point(s) for the given function, we need to analyze the first and second derivatives of the function.
Let's start by finding the first derivative, f'(x), of the function f(x) = 3x^(1/3) + 6x^(4/3).
Step 1: Find f'(x)
To find f'(x), we need to differentiate each term of the function separately using the power rule.
Differentiating the first term, 3x^(1/3):
Using the power rule, the derivative of x^n, where n is a constant, is given by nx^(n-1). Applying this rule, we have:
d/dx (3x^(1/3)) = (1/3) * 3 * x^(-2/3) = x^(-2/3)
Differentiating the second term, 6x^(4/3):
Using the power rule again, we have:
d/dx (6x^(4/3)) = (4/3) * 6 * x^(1/3) = 8x^(1/3)
Combining the derivatives, we get:
f'(x) = x^(-2/3) + 8x^(1/3)
Step 2: Analyze f'(x) for extrema
To find the x-coordinates of any relative extrema, we need to solve the equation f'(x) = 0.
Setting f'(x) = 0, we have:
x^(-2/3) + 8x^(1/3) = 0
To solve this equation, we can factor out a common term:
x^(-2/3)(1 + 8x) = 0
Setting each factor equal to zero, we get:
x^(-2/3) = 0 -> x cannot be equal to zero since it would make the function undefined.
1 + 8x = 0 -> 8x = -1 -> x = -1/8
Therefore, we found one x-coordinate for the relative extrema, which is x = -1/8.
Step 3: Find the second derivative, f''(x)
To check if this x-coordinate (-1/8) corresponds to a relative maximum or minimum, we need to find the second derivative, f''(x), of the function.
Differentiating f'(x) = x^(-2/3) + 8x^(1/3) using the power rule, we have:
f''(x) = d/dx [x^(-2/3) + 8x^(1/3)]
= -2/3 * x^(-2/3 - 1) + 8/3 * x^(1/3 - 1)
= -2/3 * x^(-5/3) + 8/3 * x^(-2/3)
Step 4: Analyze f''(x) for inflection point(s)
To find the x-coordinate(s) of any inflection point(s), we need to solve the equation f''(x) = 0.
Setting f''(x) = 0, we have:
-2/3 * x^(-5/3) + 8/3 * x^(-2/3) = 0
We can simplify this equation by multiplying through by -3/2:
x^(-5/3) - 4x^(-2/3) = 0
Let's factor out a common term to solve for the x-coordinate(s):
x^(-2/3)(x^(-1) - 4) = 0
Setting each factor equal to zero, we get two possible x-coordinates:
x^(-2/3) = 0 -> x cannot be equal to zero since it would make the function undefined.
x^(-1) - 4 = 0 -> x^(-1) = 4 -> 1/x = 4 -> x = 1/4
Therefore, we found one x-coordinate for the inflection point, which is x = 1/4.
Summary of x-coordinates:
- Relative Extrema: -1/8
- Inflection Point: 1/4
To fully justify the answer, you can also conduct further analysis by applying the First or Second Derivative Test, depending on the nature of the extrema and inflection point.