Question

+ Find the greatest value for the expression (a+x) ^3(a-x)^4 for x<a​

Answer 1

Answer:

To find greatest value of;

[(a+x)^3 ][(a-x)^4]=f (x) (say)

We are to find the point(s) of extremum of the given function. (the differentiation method is appropriate here)

To determine which point maximizes the function.

Given; x<a => (a-x)>0 ( log function can be imposed upon the given function for ease of calculation)

So; taking log on given function we get;

log (f (x))= 3log (a+x)+4log (a-x)=g (x) (let)

Now; differentiating f (x) and finding the extreme points by equating to 0 is equivalent to differentiating log (f (x)) and finding the extreme points, as log(.) function is monotonic.

So; d/dx (log (f (x))=0

Or, d/dx (3log (a+x)+4log (a-x))=0

Or,[ 3/(a+x)]-[4/(a-x)]=0

Or, 3/(a+x)=4/(a-x)

Or, (a-x)/(a+x)=4/3

Or,2a/-2x=7/1 (by componendo and dividendo)

Or, x=-a/7

Now to check if the extremun point x=-a/7 is maxima or not;double differentiation log (f (x))

g’’(x)[at x=-a/7]

=[-3/(a+x)^2]+[4 /(a-x)^2] (at x=-a/7)

=[(7/2)^2][(-1/12)+(1/8)] < 0

So according to rule maxima exists at ;

x=-a/7

The required maximum is ;

[(a-7/a)^3][(a+7/a)^4]

=[(2a/7)^7][12^3×4]

=1.074a