What are the critical points of h(x)= 5x+(4/x) and are they minimum or maxmum values.

Expert Answers
justaguide eNotes educator| Certified Educator

We have the function h(x) = 5x + (4/x). The critical points of this function are at points where h'(x) = 0

h'(x) = 5 - 4/x^2

5 - 4/x^2 = 0

=> x^2 = 4/5

=> x= 2/sqrt 5 and -2/sqrt 5

h''(x) = 8/x^3

At x = 2/sqrt 5, h''(x) is positive, therefore we have a local minimum here.

h(2/sqrt 5) = 5*(2/ sqrt 5) + 4*sqrt 5 / 2 = 4*sqrt 5

At x = -2/sqrt 5, h''(x) is negative, therefore we have a local minimum here.

h(2/sqrt 5) = 5*(2/ sqrt 5) + 4*sqrt 5 / 2 = -4*sqrt 5

Here, we see that the local minimum has a larger value than the local maximum, this is not an error. As the function does not have an absolute maximum or an absolute minimum, the local extreme values can have the values we have obtained.

The required point of local minimum is ( 2/ sqrt 5 , 4*sqrt 5) and the point of local maximum is (-2/sqrt 5 , -4*sqrt 5)

giorgiana1976 | Student

The critical points are the roots of the 1st derivative of the function.

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

f'(x)=0

5 -4/x^2=0

5x^2 - 4 = 0

5x^2=4

x1=2/sqrt5 and x2=-2/sqrt5

we'll calculate f(2/sqrt5)=10/sqrt5+4sqrt5/2

f(2/sqrt5)=20/2sqrt5 + 20/2sqrt5=20/sqrt5

f(-2/sqrt5)=-10/sqrt5-4sqrt5/2

f(-2/sqrt5)=-20/2sqrt5-20/2sqrt5

f(-2/sqrt5)=-20/sqrt5

The function has a maximum local point for f(-2/sqrt5)=-20/sqrt5 and a minimum local point for f(2/sqrt5)=20/sqrt5.

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