Critical Numbers, Inflection Points, and y Values

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Jason76

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You find critical numbers (extrema) by setting the 1st derivative = 0, and then finding the x values. You determine whether they are maximum or minimum by plugging in critical numbers (respectively) into the 2nd derivative(while observing whether they are + or -). But how do you find the y values of critical number x values?

Likewise, you find the inflection points by setting the 2nd derivative = 0, and solving for x values, but how do you find the y values which correspond?
 
Last edited:
You plug the x-values into the original function.
 
Jason76 said:
You find critical numbers (extrema) by setting the 1st derivative = 0, and then finding the x values. You determine whether they are maximum or minimum by plugging in critical numbers (respectively) into the 2nd derivative. But how do you find the y values of critical number x values?

Likewise, you find the inflection points by setting the 2nd derivative = 0, and solving for x values, but how do you find the y values which correspond?
Click to expand...

All the points look like \(\displaystyle \left( {{x},f\left( {{x}} \right)} \right).\) Thus \(\displaystyle y={f\left( {{x_0}} \right)}\)
Where \(\displaystyle x_0\) is the critical value.
 
Jason76 said:
You find critical numbers (extrema) by setting the 1st derivative = 0, and then finding the x values.

You determine whether they are maximum or minimum by plugging in critical numbers (respectively) into the 2nd derivative
(while observing whether they are + or -).
Click to expand...

You don't have to use the 2nd derivative to determine if there is a relative maximum or relative minimum at a critical number.

If the function is not undefined at a certain critical number, then you can use appropriate choices of test numbers on either
side of the critical number. Look at the signs of the slopes at the test numbers. Look at the following cases of the orders
of the signs of the slopes ** at the test numbers:


1) increase, decrease ----> relative maximum

2) decrease, increase ----> relative minimum

3) increase, increase or decrease, decrease ----> neither relative maximum nor relative minimum




**

+ slope at a critical number ----> interval of increase

- slope at a critical number ----> interval of decrease


/\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\../\


Edit:

Likewise, you find the inflection points by setting the 2nd derivative = 0, and solving for x values, ...[/QUOTE]

There may or there may not be inflection points at the x-values you worked out from the 2nd derivative equal to zero.

One of the ways to make sure is to use test points. If the critical number from the second derivative occurs at where
the function is not undefined, then the sign of the second derivatives evaluated at consecutive second derivative
test numbers will have to change from + to -, or vice versa.
 
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lookagain said:
You don't have to use the 2nd derivative to determine if there is a relative maximum or relative minimum at a critical number.

If the function is not undefined at a certain critical number, then you can use appropriate choices of test numbers on either
side of the critical number. Look at the signs of the slopes at the test numbers. Look at the following cases of the orders
of the signs of the slopes ** at the test numbers:


1) increase, decrease ----> relative maximum

2) decrease, increase ----> relative minimum

3) increase, increase or decrease, decrease ----> neither relative maximum nor relative minimum




**

+ slope at a critical number ----> interval of increase

- slope at a critical number ----> interval of decrease
Click to expand...

This is definitely the way to go when the second derivative is a BEAST to do.
 
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