max and min coordinates

[Mel Mitch]

hello everyonE,
This should be simple.....can someone help

Problem

y= x^3 + 2x^2 - 4x - 4

find the max and min coordinates

 

[galactus]

Differentiate, set to 0 and solve for x.

Once you differentiate, you should get a quadratic which is easy to solve.

 

[Deleted member 4993]

hello everyonE,
This should be simple.....can someone help

Problem

y= x^3 + 2x^2 - 4x - 4

find the max and min coordinates

Yes - it is if you know a bit of calculus. I am going to assume that you know for min/max f'(x) = 0 and for max f"(x)<0 and for min f"(x)>0.

Lets assume your function is:

f(x) = 6*x[sup:3r04unx9]3[/sup:3r04unx9] + 7/2*x[sup:3r04unx9]2[/sup:3r04unx9] -12*x + 3

f'(x) = 12*x2 + 7x - 12 = (4*x-3)(3x+4) ..............f'(3/4) = 0 and f'(-4/3) = 0

f"(x) = 24*x + 7 ..............f"(3/4) = 25 >0 hence minimum ...............f"(-4/3) = -25 <0 hence maximum

f(3/4) = -1.5

f(-4/3) = 7 + 4/9 = 7.44

the function has local minimum at (0.75,-1.5) and local maximum at (-4/3, 7.44)

The answers should be verified by using a calculator.

 

[Mel Mitch]

thanks Subhotosh Khan that was very helpful....will be able to work out my problems...thanks again