Thread: Extreme Points

Determine the extreme points
F(x)= 3x^4 + 16x^3 + 24x^2 + 32

Any help is appreciated. Thanks!

Originally Posted by akuczma86

Determine the extreme points
F(x)= 3x^4 + 16x^3 + 24x^2 + 32

Any help is appreciated. Thanks!

Hi akuczma86,

all you need do is differentiate the function with respect to x.
This gives an equation for the slope of the tangent to the curve for all x.
The extreme points occur where the slope of the tangent is zero (the tangent is horizontal).

Hence









which means at the local extremes,

Factorising this gives the values of x where the local extremes occur.





The possible extremes occur when x=0 and x+2=0

hence and

Now substitute these x into F(x) to find the vertical co-ordinates.

The 2nd derivative tells us what these possible extremes look like,
whether a local max, local min, or saddle

F''(x)>0 is a minimum
F''(x)<0 is a maximum
F''(x)=0 is a saddle point



For x=0, this is 48, so there is a local minimum at x=0

For x=-2, this is 0, which means there is a saddle point at x=-2

hence there is one extreme local minimum at (0,32)