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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beckden | High School Teacher | (Level 1) Educator

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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 to the left of x=1 and negative to the right of x=1 so x=1 is a maximum. (slope is decreasing around x=1).

Since there are no other critical points, x=1 is an absolute maximum, and f(x) is increasing from (-oo, 1),

Now we need to know when f(x) is increasing the most rapidly.  This would be when f''(x) = 0.

f''(x) = (d)/(dx) 4(1-x)/(x^2+4)^(-3/2)

f''(x) = 4(1-x)(d)/(dx)(x^2+4)^(-3/2) + 4(x^2+4)^(-3/2)(d)/(dx)(1-x)

f''(x) = 4(1-x)(-3/2)(x^2+4)^(-5/2)(2x) + 4(x^2+4)^(-3/2)(-1)

f''(x) = -6(1-x)x(x^2+4)^(-5/2) - 4(x^2+4)(x^2+4)^(-5/2)

f''(x) = ((6x^2+6)-(4x^2+4))(x^2+4)^(-5/2)=(2x^2-2)(x^2+4)^(-5/2)

This is zero when (2x^2-2)=0 or x=-1, +1.

f''(x) is positive when x<-1, negative when -1<x<1, and positive when x>1,

x=-1 is maximum for f'(x) and x=1 is a minimum for f'(x).

So the answer is x=1 is an absolute maximum, and at x=1 f(x) f is increasing the most rapidly.

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