30 is your mystery answer :)
The local minimum of function is an argument x for which the first derivative of function g(x) is equal to zero, so:
g'(x)=0
g'(x)=(x^4-5x^2+4)'=4x^3-10x=0
x(4x^2-10)=0
x=0 or 4x^2-10=0
4x^2-10=0 /4
x^2-10/4=0
x^2-5/2=0
[x-sqrt(5/2)][x+sqrt(5/2)]=0
Now we have to check wchich argument gives the minimum value from x=0, x=sqrt(5/2) and x=-sqrt(5/2).
g(0)=4
g(sqrt(5/2))=25/4-5*5/2+4=4-25/4=-9/4
g(-sqrt(5/2))=-9/4
The answer is sqrt(5/2) and -sqrt(5/2).
Answer:
atm fees that is your answer
Step-by-step explanation:
Solution:
100<em>x</em>^2-58<em>x</em>-16
