find any stationary points of the function g(x) = (2x-3)square root of 5+16x-4x^2 (0 less than or equal to x less than or equal to 4) and use the first derivative test to classify each stationary point as a local maximum or local minimum of g(x)
this is how far i have got with an answer
g'(x) = (2x-3) (sqrt(5+16x-4x^2) + 2sqrt(5+16x-4x^2)
= (-4x+16)(2x-2)/sqrt(5+16x-4x^2) + 2sqrt(5+16x-4x^2)
Simplify:
-2(8x^2-30x+7) / sqrt(-4x^2+16x+5)
Set this to 0:
0 = -2(8x^2-30x+7) / sqrt(-4x^2+16x+5)
0 = -2(8x^2-30x+7)
Factor:
0 = -2(2x-7)(4x-1)
Set each factor to 0:
2x-7 = 0, so x = 7/2
4x-1 = 0, so x = 1/4
can't work out how to get the xLEFT and xRIGHT to classify each using the first derivative to get local min or local max...
help!
Use the second derivative. It it is positive at the stationary point the function is about to get bigger so that is a minimum. If it is negative, the function is about to get smaller so that is a maximum.
By the way I did not check what you did. I suspect it is right because you came out with reasonable points where the function is horizontal.
call your tutor rather than trying to get someone else to do your course work for you
To classify each stationary point as either a local maximum or local minimum, we need to analyze the behavior of the function around these points using the first derivative test.
First, we need to find the critical points of the function by setting the first derivative equal to zero. From your calculations, we have:
g'(x) = -2(8x^2-30x+7) / sqrt(-4x^2+16x+5)
Setting this to zero, we have:
0 = -2(8x^2-30x+7)
Now, we can solve for x:
2x^2 - 15x + 7 = 0
This quadratic equation does not factor nicely, so we can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 2, b = -15, and c = 7. Solving this equation, we will get two values for x, which correspond to the critical points of the function.
Once we have the critical points, we can use the first derivative test to classify them as local maximum or local minimum. The first derivative test involves evaluating the sign of the derivative to the left and right of each critical point.
To determine the signs, we can choose test points within each interval and evaluate the sign of the derivative at those points.
For example, let's take x = 0 as a test point for the interval to the left of the critical point. Evaluate g'(0) to find the sign. If g'(0) is positive, then the function is increasing to the left of the critical point, indicating a local minimum. If g'(0) is negative, then the function is decreasing to the left of the critical point, indicating a local maximum.
Repeat the same process for a test point to the right of the critical point. If the sign changes from positive to negative, we have a local maximum. If the sign changes from negative to positive, we have a local minimum.
I hope this helps you classify the stationary points of the function g(x) and determine whether they are local maxima or minima.