help for max min problem, is this right?

Question

problem:
find absolute madimum and minimum values, if they exist, over the indicated interval
the equation is
f(x) 3/4x^4 + 1/3x^3 -3x; [-1,3]

I got no max, and min of .90033

is that right?

Answer 1

Actually, you have an endpoint maximum ...

since you gave a closed interval of [ -1, 3 ] ,
you need to look for critical points and check the end points ....and compare their y values .... and x = 3 gives a max of y = 60.75

If you meant ( -1, 3) or [ -1, 3) then
no Max at x = 3

now for the min... I got x = 0.9003 and
y = -1.965 on TI calculator....

by derivatives... 3x^3 + x^2 - 3 = 0 is not easy to solve ....but again, it has a zero at
x = 0.9003. There is an eqaution for solving cubics, but I don't remember it at the moment...

y '' = 9x^2 + 2x, and at x = 0.90033 , y '' is > 0, indicating a minimum....

CPUcate... the problem you solved is not the one given above... sorry, your answer is incorrect....

Answer 2

each right this moment piece is a radius of the circle. assume the attitude between them is x radians and the radius is x. The circumference of the great circle is two*pi*r The length of the arc is (x/(2*pi))*2*pi*r = xr So the entire length of the fence is: xr + 2r = 10 xr = 10 - 2r x = (10 - 2r) / r x = 10r^(-a million) - 2 the area of the great circle is pi*r^2 the area enclosed with the aid of the fence is: A = pi * r^2 * (x/(2*pi)) A = r^2 * (x/2) A = r^2 * (10r^(-a million) - 2) / 2 A = r^2 * (5r^(-a million) - a million) A = 5r - r^2 dA/dr = 5 - 2r to maximise the area, set the spinoff to 0: 5 - 2r = 0 2r = 5 r = 5/2 r = 2.5 x = (10 - 2*2.5) / 2.5 x = (10 - 5) / 2.5 x = 5 / 2.5 x = 2 The length of the arc is x*r = 2 * 2.5 = 5

Answer 3

f(x) = 3/4 x^3 - 3x
f '(x) = 9/4 x^2 - 3 = 0
9/4 x^2 = 3
x^2 = 4/3
x = 2/sqr(3) = 2 sqr(3) /3

Answer 4

f(x) = 3/4x^4 + 1/3x^3 - 3x; [-1,3]
f'(x) = 3x^3 + x^2 - 3; [-1,3]
f(0.900329) = - 1.964925495 (min)
f(3) = 60.75 (max)