Calculus: Identify the open intervals..?

Question

Identify the open intervals on which the function is increasing or decreasing

y = x + (4/x)

Answer 1

To do this question using calculus, you must take the first derivative of your function. Let's do this step by step:

First, the equation:
y = x + (4/x)

let's convert it to a format more easily derived:
y = x^1 + 4x^(-1)

Notice that this is the same as the original eqation, just in a modified form.

Now, the derivation;
When deriving, rember that y' = nx^(n-1), where n is the initial exponent. also remember that the derivative of a sum is the same as the sum of its derivatives.

Using the above rules, we find that;

the derivative of x^1 is 1x^(1-1), but anything raised to the zero power is just 1.

the derivative of 4x^(-1) is 4((-1)x^(-1-1).
Simplified, this becomes -4x^-2, or -4/(x^2).

so we have now derived the equation,

y' = 1 - (4/(x^2))

The next thing to remember is that to identify the intervals, you need to find the critical values of the equation. this would be where the solution is either equal to zero or is undefined. In this case, x cannot equal 0, there it would be undefined; also the values 2 and -2 result in a solution of 0. These will be the points we will focus on to determine the intervals.

to determine whether it is increasing or decreasing, we will us the sign of the resulting y value of the equation. if it is negative, the function is decreasing, if it is positive, the function is increasing over the interval.

<___-2___0___2___>

that is our number line, now pick a value for x which is less than -2 and substitute it into the function y':

x = -10

y' = 1 - (4/(-10^2))

y' = 1 - (4/100)

y' = 24/25 <--------- positive value

Now a number between -2 and 0:

x = -1

y' = 1 - (4/(-1^2))

y' = 1 - (4/1)

y' = -3 <------------- negative value

now between 0 and 2:

x = 1

y' = 1 - (4/(1^2))

y' = 1 - (4/1)

y' = -3 <------------- negative value

Finally, from -2 to infinity:

x = 10

y' = 1 - (4/(10^2))

y' = 1 - (4/100)

y' = 24/25 <--------------- positive value

Check the nubmer line

<__+__-2__-__0__-__2__+__>

showing that the function changes signs over -2 and 2, we now have the solution, you just have to interpret it.

When y' is negative, the original function is decreasing, when it is positive, the function is increasing, therefore:

The funtion is increasing over the open interval:

( (-infinity) --- -2 ) U ( 2 --- (infinity) )

and decreasing over the open interval:

( -2 --- 0 ) U ( 0 --- 2 )

Answer 2

y = x + (4/x)
y' = 1 - 4/x^2
for 0 slope,
x^2 = 4
x = ± 2
increasing when x < - 2
decreasing when - 2 < x < 0
decreasing when 0 < x < 2
increasing when 2 < x

Answer 3

It's been a while since I took Calc but I think you take the derivative, graph it and where it is negative, it's decreasing and where its positive, it's increasing.