Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)=9x^1/3+9/2x^4/3.
I've gotten -0.5 as the relative minimum and x=1 for the inflection points. Is this correct? If not, can you show me how to get the correct answer please
Your lack of brackets leave the interpretation of the equation very ambiguous.
e.g. do you mean: f(x)=9x^(1/3) + (9/2)x^(4/3) ?
please retype before I attempt it
yes I mean f(x)=9x^(1/3) + (9/2)x^(4/3). Sorry, I've been trying to figure this out for hours
Looks pretty straight forward ...
f(x)=9x^(1/3) + (9/2)x^(4/3)
f ' (x) = 3x^(-2/3) + 6x^(1/3)
f''(x) = -2x^(-5/3) + 2x^(-2/3)
for relative max/min, f'(x) = 0
3x^(-2/3) + 6x^(1/3) = 0
3x^(-2/3) (1 + 2x) = 0
3x^(-2/3) = 0 ---> no solution
or
1+2x = 0
x = -1/2 or -.5 ------- you had that
for points of inflection,
-2x^(-5/3) + 2x^(-2/3) = 0
-2x^(-5/3) (1 + x) = 0
x = -1
I agree with both of your answers
To find the x-coordinates of the relative extrema and inflection points for the given function f(x) = 9x^(1/3) + (9/2)x^(4/3), we need to take the first and second derivatives of the function.
Step 1: Find the first derivative f'(x):
To find the first derivative, we use the power rule. For each term, we multiply the coefficient by the exponent and subtract 1 from the exponent.
For f(x) = 9x^(1/3) + (9/2)x^(4/3), the first derivative is:
f'(x) = (1/3) * 9x^(-2/3) + (4/3) * (9/2) * x^(1/3)
Simplifying f'(x):
f'(x) = 3x^(-2/3) + 6x^(1/3)
Step 2: Find the second derivative f''(x):
To find the second derivative, we differentiate the first derivative with respect to x.
For f'(x) = 3x^(-2/3) + 6x^(1/3), the second derivative is:
f''(x) = (-2/3) * 3x^(-5/3) + (1/3) * 6x^(-2/3)
Simplifying f''(x):
f''(x) = -2x^(-5/3) + 2x^(-2/3)
To find the relative extrema, we set the first derivative f'(x) equal to zero and solve for x:
0 = 3x^(-2/3) + 6x^(1/3)
Dividing by 3:
0 = x^(-2/3) + 2x^(1/3)
Now, we can simplify using a common denominator and solve for x:
0 = x^(-2/3) + 2x^(1/3)
0 = (x + 2x^(3/3)) / x^(2/3)
0 = (x + 2x) / x^(2/3)
0 = 3x / x^(2/3)
0 = 3 / x^(1/3)
To solve for x:
x^(1/3) = 3
Raise both sides to the power of 3:
x = 3^3
x = 27
So, we have one critical point x = 27.
To determine if it is a relative minimum or maximum, we need to use the second derivative test. We substitute the critical point x = 27 into the second derivative f''(x):
f''(27) = -2(27)^(-5/3) + 2(27)^(-2/3)
f''(27) = -2(1/27^(5/3)) + 2(1/27^(2/3))
f''(27) = -2/27^(5/3) + 2/27^(2/3)
f''(27) = -2/27^(5/3) + 2/27^(2/3)
Calculating this value, we find that f''(27) is negative. Therefore, x = 27 is the x-coordinate of a relative maximum.
To find the inflection point(s), we set the second derivative f''(x) equal to zero and solve for x:
0 = -2x^(-5/3) + 2x^(-2/3)
Dividing by 2:
0 = -x^(-5/3) + x^(-2/3)
Adding x^(-5/3) to both sides to isolate x^(-2/3):
x^(-5/3) = x^(-2/3)
Raise both sides to the power of -3/5:
(x^(-5/3))^(-3/5) = (x^(-2/3))^(-3/5)
x = x^(-6/5)
1 = x^(-6/5)
To solve for x:
x^(6/5) = 1
Raise both sides to the power of 5/6:
x = 1^(5/6)
x = 1
So, we have one inflection point x = 1.
In conclusion, the x-coordinates of the relative extrema for the given function f(x) = 9x^(1/3) + (9/2)x^(4/3) are x = 27 (relative maximum) and the x-coordinate of the inflection point is x = 1.