Please help!!
*f(x)= x(x+6)^1/2 find two x intercepts.
Then show that f'(x)=0 at some point between the 2 x intercepts.
*Use mean value theorem for f(x)= x(x^2-3x+1) for interval [1,4]
I FOUND DERIAVITIVE WHICH IS 3X^2-6X+1.THEN?
*find all critical numbers find all intervals on which the function is increasing or decreasing. lable extrma.
I FOUND MY DERIVATIVE WHICH IS -32X/(X^2-16)^2
CRITICAL NUMBERS ARE 0,4 AND -4.
SO WHAT DO i DO NEXT
the two x intercepts are 0, and -6
set the derivative equal to zero, and find x
f'(x)=(x+6)^1/2 + x/2 *1/(x+6)^2=0
x+6 + x/2=0
3x=-12
x=-4
***
find f(1), and f(4)
f(1)=-1
f(4)=20
this means there is a zero between x=1,4
*****
I wonder on this third question what is the function? Putting parts of several qustion on one question is very confusing.
f(x)=x^2/x^2-16
i got this. thnks
To find the x-intercepts of the function f(x) = x(x+6)^(1/2), we need to set the function equal to zero and solve for x. In other words, we want to find the values of x for which f(x) = 0.
Setting x(x+6)^(1/2) = 0, we have two cases:
1. x = 0
2. (x+6)^(1/2) = 0
In the second case, the square root of a number can never be equal to 0. So, we only have one x-intercept at x = 0.
To show that f'(x) = 0 at some point between the two x-intercepts, we need to find the derivative of f(x) and determine where it equals zero.
To find the derivative of f(x) = x(x+6)^(1/2), we can use the product rule and the chain rule.
The derivative is given by f'(x) = (x+6)^(1/2) + (x/2)(x+6)^(-1/2).
Next, we need to find the points where f'(x) = 0. So, we set f'(x) = 0 and solve for x.
(x+6)^(1/2) + (x/2)(x+6)^(-1/2) = 0
Simplifying this equation can be a bit tedious, but you can start by multiplying both sides by (x+6)^(1/2) to get rid of the denominators. Rearrange the terms and solve for x to find the values where f'(x) = 0.
Moving on to your next question, regarding the mean value theorem for the function f(x) = x(x^2 - 3x + 1) on the interval [1, 4].
To apply the mean value theorem, we need to verify that the function f(x) is continuous on the closed interval [1, 4] and differentiable on the open interval (1, 4).
Since f(x) is a polynomial function, it is continuous and differentiable for all real numbers. So, it satisfies the requirements of the mean value theorem.
To use the mean value theorem, we need to find the average rate of change of f(x) over the interval [1, 4]. The average rate of change is given by (f(b) - f(a))/(b - a), where a and b are the endpoints of the interval.
Evaluate f(4) and f(1), then calculate the difference and divide it by the width of the interval (4 - 1).
After obtaining the average rate of change, we can apply the mean value theorem. It states that for some value c in the open interval (1, 4), the derivative of f(x) evaluated at c is equal to the average rate of change obtained earlier.
Finally, you mentioned that you found the derivative of f(x) = x(x^2 - 3x + 1) to be 3x^2 - 6x + 1. If that's the case, you can proceed with the steps described above using the derivative you found.
Moving on to the last part, where you need to find all critical numbers, intervals of increase/decrease, and label extrema for the function.
You mentioned that the derivative you found is -32x/(x^2 - 16)^2 and that the critical numbers are 0, 4, and -4.
To determine the intervals of increase and decrease, you can use the critical numbers and test points in each interval.
An interval is increasing if the derivative is positive in that interval, and it is decreasing if the derivative is negative.
Evaluate the derivative at test points in each interval: a point smaller than -4, a point between -4 and 0, a point between 0 and 4, and a point greater than 4.
If the derivative is positive at a test point, then that interval is increasing. If the derivative is negative at a test point, then that interval is decreasing.
To determine extrema, check the sign changes of the derivative. If the sign changes from positive to negative, there is a local maximum. If the sign changes from negative to positive, there is a local minimum.
By analyzing the intervals and the sign of the derivative, you can determine where the function is increasing, where it is decreasing, and locate any extrema on those intervals.
I hope this helps you solve your problem step by step! If you have any further questions, feel free to ask.