f(x)=(x+4)/(sqrt(x^2+4) Find the exact value of the maximum of f and Find the exact value of x at which f increases most rapidly.
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f'(x)=(x+4)/sqrt(x^2+4)
f'(x)=(x+4)(d)/(dx)((x^2+4)^(-1/2))+(x^2+4)^(-1/2)(d(x+4))/(dx)
f'(x)=(x+4)(-1/2)(x^2+4)^(-3/2)(2x)+(x^2+4)^(-1/2)
f'(x)=-((x+4)x)(x^2+4)^(-3/2)+(x^2+4)(x^2+4)^(-3/2)
f'(x)=(4-4x))/(x^2+4)^(3/2)
f'(x)=0 when (4-4x)=0 x=1
Now we need to find if this is a minimum or maximum.
Since (x^2+4)>0 always we need to find out the signs of (4-4x)
(4-4x) > 0 when x < 1 and (4-4x)<0 when x>1
So the derivative is positive...
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