What is the maximum value of the function f(x)=A*exp(-a*x^2-b*x-c) where A,a,b,c are positive constants?

Answer 1

the above is maximum when ax^2+bx+c is minimum

ax^2+bx+c is minium or c when x = 0 as a,b are positive

so maximum value f(x) is at 0 that is A exp(-c)

Answer 2

Since A is a constant only the exponents matter y, so take the first derivative which is d y/dx= -2ax-b and equate it to 0.
ie: -2ax-b=0, when a, b are known we can find the value of x which when substituted in the given f(x) to get the maximum value of it. Here the second derivative is negative meaning there is positive maximum at x got above. Of course A should also be know along with a,b and c.

Answer 3

SP is loopy... the respond is 0. no count what x is, once you subtract x from it, you get 0. something expanded with 0 provides 0. (x-a)(x-b)...(x-x)...(x-z)..=0 on the grounds which you have reported a=one million,b=2,c=3,....z=26... x could desire to be 24.

Answer 4

f(x)=A*exp(-a*x^2-b*x-c)
diff wrt x
f'(x)=A*[-2ax-b]*exp(-a*x^2-b*x-c)....(1)
put f'(x)=0
since exp(-a*x^2-b*x-c)cannot be zero
so[-2ax-b]=0
x=-b/2a
again diff (1)
f'(x)=A*{[-2ax-b]^2*exp(-a*x^2-b*x-c)
+(-2a)exp(-a*x^2-b*x-c)}
put x=-b/2a u get negative term
so x=-b/2a as point of maxima
so max. value of f(x)is
=A*exp(-a*(-b/2a)^2-b*(-b/2a)-c)
solve it

Answer 5

you got it in one dupinder

f(x)max=Ae^((b^2/4a)-c)

Answer 6

YES!