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
$\displaystyle F(x)=3x^4+16x^3+24x^2+32$
$\displaystyle F'(x)=4\left(3x^3\right)+3\left(16x^2\right)+2\lef t(24x\right)+0=0$
$\displaystyle =12x^3+48x^2+48x=0$
$\displaystyle =12\left(x^3+4x^2+4x\right)=0$
which means at the local extremes, $\displaystyle x^3+4x^2+4x=0$
Factorising this gives the values of x where the local extremes occur.
$\displaystyle x\left(x^2+4x+4\right)=0$
$\displaystyle x(x+2)^2=0$
The possible extremes occur when x=0 and x+2=0
hence $\displaystyle x=0$ and $\displaystyle x=-2$
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
$\displaystyle F''(x)=36x^2+96x+48$
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)