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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)=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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