Suppose a normal distribution has a mean of 16 and a standard deviation of 4.
2 answers:
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The graph of <span>y=-0.5 sqrt (x-3)+2
Df= {x/x-3>=0}
Df= [3, + infinity[
derivative of f(x)
f'(x)= -0.5 x 2 /</span>sqrt (x-3)= - 1/sqrt (x-3) <0, f is a decreasing function for all x in the Df
limf(x)=2 x--------->3, limf(x)=-infinity, x--------->+infinity
look at the graph
Df= {x/x-3>=0}
Df= [3, + infinity[
derivative of f(x)
f'(x)= -0.5 x 2 /</span>sqrt (x-3)= - 1/sqrt (x-3) <0, f is a decreasing function for all x in the Df
limf(x)=2 x--------->3, limf(x)=-infinity, x--------->+infinity
look at the graph