state increasing, decreasing, max, min
for f(x)= x^(1/3) (x+4)
Here is my work:
f'(x)= 4/3x^(1/3) + 4/3x^(-2/3)
4/3x^(1/3) + 4/3x^(-2/3)=0
x=0 x= -1
Increasing: [0, ∞)
Decreasing: (-∞, 0]
min @ x=0
max: none
Is this correct?
Where does the (4/3) come from in your answer?
the original was
f(x) = x^(1/3) (x+4) , do you have a typo?
using the product rule, I got
f '(x) = x^(1/3) + (x+4)(1/3)x^(-2/3)
common factor of x^(-2/3) ...
= x^(-2/3) (x + x+4)
= x^(-2/3) (2x+4)
or
(2x+4)/x^(2/3) = (2x+4)/(x^2)^(1/3)
the denominator is always positive
max/min ? when 2x+4 = 0
x = -2, the f(x) = 2(-2)^1/3)
we can see that the curve crosses at 0 and -4
so x=-2 would produce a minimum.
There is no maximum
looking at the f '(x) of (2x+4)/(x^(2/3))
we already saw that the bottom is always positive except x=0,
so all we need to do is look at the top
2x+4 > 0 for all x> -2
and
2x+4 < 0 for all x < -2
I combined right off the bat, so
f(x)= x^(1/3) (x+4)
f(x)= x^(4/3) + 4x^(1/3)
f'(x)= (4/3)x^(1/3) + (4/3)x^(-2/3)
(4/3)x^(1/3) + (4/3)x^(-2/3)=0
x=0 x= -1
...I can't find a flaw in my calculations..
Ok, I see now,
so your derivative is ok
but your value of x=0 doesn't satisfy
(4/3)x^(1/3) + (4/3)x^(-2/3)=0
let x = 0
LS = (4/3)(0) + (4/3)(1/0) which is undefined
(4/3)x^(1/3) + (4/3)x^(-2/3)=0
divide both terms by 4/3
x^(1/3) + x^(-2/3)=0
common factor of x^(-2/3)
x^(-2/3) ( x + 1) = 0
x = -1
ahhh, let's check mine, looks like I forgot the 1/3 when I factored.
f '(x) = x^(1/3) + (x+4)(1/3)x^(-2/3)
common factor of x^(-2/3) ...
= x^(-2/3) (x + (1/3)(x+4))
= x^(-2/3) (3x+x+4)/3
= x^(-2/3)(4x+4)/3
which when I set it equal to zero gives me x=-1 , the same as yours
the rest of my argument is valid
so x = -1 yields a minimum value of (-1)^(1/3) (3)
= -3
increasing for x > -1
decreasing for x < -1
Okay, so I was right apart from the x=0. Thank you
also, change your answer for increasing and decreasing.
Naturally, yes
Yes, your work is mostly correct. Let's go through it step by step:
1. To determine the critical points of the function f(x), we need to find where the derivative f'(x) is equal to zero or undefined.
2. Start by finding the first derivative of f(x) using the power rule:
f'(x) = (1/3)x^(-2/3)(x+4) + x^(1/3)(1)
3. Simplify the equation:
f'(x) = (1/3)x^(-2/3)(x+4) + x^(1/3)
4. To find where the derivative is equal to zero, set f'(x) equal to zero and solve for x:
(1/3)x^(-2/3)(x+4) + x^(1/3) = 0
5. Multiply through by 3 to remove the fraction:
x^(-2/3)(x+4) + 3x^(1/3) = 0
6. Multiply through by x^(2/3) to clear the denominator:
(x+4) + 3x^(5/3) = 0
7. Combine like terms:
4 + 3x^(5/3) = 0
8. Subtract 4 from both sides:
3x^(5/3) = -4
9. Divide by 3:
x^(5/3) = -4/3
10. Take the fifth root of both sides to solve for x:
x = (-4/3)^(3/5) ≈ -0.8706
11. So one critical point is x ≈ -0.8706.
12. Another critical point occurs where the function is undefined, which is at x = 0.
Now we can analyze the intervals and determine the increasing and decreasing behavior:
- Between negative infinity and x ≈ -0.8706, the function is increasing.
- Between x ≈ -0.8706 and 0, the function is decreasing.
- After x = 0, the function is increasing.
So, your intervals for increasing and decreasing are correct. However, there is no maximum or minimum since the function does not have any local extrema. Therefore, your conclusion is correct.