How do you find the maximum point from y= 24^2 - 192x +12x^3?

Answer 1

Maximum points occur at points where the slope of the tangent line is zero ***OR AT THE ENDPOINTS** if you're dealing with a segment.

1. Take the derivative of the function
2. Set the the derivative to zero
3. Solve the equation in #2 to find *possible* maximum/minimum.
4. Plug x - value(s) found in #3 into the ORIGINAL equation
5. Choose the x values in #4 that produce the largest y -- there could be a tie. . .

y' = -192 + 36 x ^2
0 = -192 + 36 x ^2
x = +/- sqrt(192/36)

y = approx 280 at positive sqrt and approx 871 at negative sqrt

So, max occurs at x = - sqrt(192/36)

I hope my computations are correct -- the steps to finding maxima and minima is correct, though.

Answer 2

you would need to find the first and second derivative to find the maximum point

y = 24x^2 - 192x + 12x^3

y' = 48x - 192 + 36x^2

y'' = 48 + 72x

all stationary points will yeild y' = 0

ie 36x^2 + 48x - 192 = 0
=> 12 (3x^2 + 4x - 16) = 0
=> 12 (3x )(x ) = 0

use that above to find the stationary points then plug these values into the equation for y'' and determine whether it is zero, positive or negative.

for a maximum point y'' would be negative

Answer 3

y' = 36x^2 - 192
y' = 0 when x = +/- (4/3)sqrt(3)
max occurs when x = - (4/3)sqrt(3)
min occurs when x = (4/3)sqrt(3)