Calculus question? Find the critical numbers of f (x) = x^2-6x. Find also the open intervals on which the function is increasing or decreasing and locate all relative extrema.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To establish the monotony of a function,only the first derivative of the function is useful. So, if the first derivative is positive on an interval, then the function id increasing on that interval and if the first derivative is negative on an interval, then the function is decreasing on that interval.

Let's find the first derivative of the function f(x):

f'(x)=(x^2-6x)'=(x^2)'-(6x)'

f'(x)=2x-6

To find out the intervals where the first derivative is positive or negative, we have to find out first, where the first derivative is annuling.For this reason we have to calculate the equation of the first derivative:

2x-6=0

2x=6

x=6/2

x=3

To discuss if the first derivative is positive or negative, we can discuss it as a linear function, knowing the rule that claims that the values of the expresion, from the left side of the root of the first derivative, have the opposite sign of the coefficient of x, which is positive, the coefficient being =2.

So, the function is decreasing on the interval (-inf, 3), because the first derivative is negative on this interval.

The values of the expresion, from the right side of the root of the first derivative, have the same sign of the coefficient of x, that means that is positive, the coefficient being =2.

So, the function is increasing on the interval (3, +inf), because the first derivative is positive on this interval.

That means that the function has an extreme point, of minimum, where the first derivative is canceling.

So, the minimum point has the coordinates (3,f(3)), where

f(3)=3^2-6*3=9-18=-9.

So, the local extreme point has the coordinates (3,-9).

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = x^2-6x.

f'(x) = 2x-6.

Equating 2x-6 = 0 we get 2x=3 or x=3/2 is the only one critical point

Also f''(x) = 2 which is positive.

At x= 3/2 , f(3/2) = (3/2)^2-6(3/2) = 2.25-9 = -6.75 is the minimum.

For x<=3/2, f'(x) = 2x-6 < = 0 So f(x) decreasing in ] -inf ,3/2)

For x >= 3/2 , f'(x) = 2x-6 > 0. So f(x) is increasing in (3/2, infinity[.

The function f(x)

 

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