Finding Critical Points Part I

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Hckyplayer8

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Find the critical points for f(x) = x2 - 7x+17
Determine the intervals where f is increasing or decreasing.
Classify each critical point as local max, local min or neither.

f'(x) = 2x-7 and set to zero x = 7/2

Completing the quadratic got me a negative square root and it is tripping me up. [7+radical -19] over [2]
 
You don't need to find when x2-7x+17=0.
Your solution "[7+radical -19] over [2] is correct (except for a +/- sign) and means there are no x-intercepts.
But this is not what the question is asking.

You have correctly solved f '(x) = 0 to get x = 7/2.
This means that there is either a local max, local min, or point of horizontal inflection when x= 7/2.
Your job is to now determine which one of these occurs.

A local minimum occurs when the gradient just to the left of the critical point is pos, and the gradient just to the right of the point is neg.
A local maximum occurs when the gradient just to the left of the critical point is neg, and the gradient just to the right of the point is pos.
If the gradient is pos on both sides or neg on both sides then you have a point of horizontal inflection (PHI).

Consider a value close to but less than x =7/2. Lets say x=3. What is f ' (3)? (you are interested in the sign here rather than the value)
Consider a point close to but more than x = 7/2/ Let's say x = 4. What is f ' (4)? (ditto)
 
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Hckyplayer8 said:
Find the critical points for f(x) = x2 - 7x+17
Determine the intervals where f is increasing or decreasing.
Classify each critical point as local max, local min or neither.

f'(x) = 2x-7 and set to zero x = 7/2

Completing the quadratic got me a negative square root and it is tripping me up. [7+radical -19] over [2]
Click to expand...
Why are you trying to do that?

To determine:

the intervals where f is increasing or decreasing.

First calculate the x for critical point - There is only one critical point. You have done that.

For these types of problems - now I would plot the function - paying special attention to the values of the function "around the critical point".

What do you see?
 
Are critical points and critical numbers interchangeable?

My previous lesson regarding critical numbers and max/mins followed the process

1. Find the derivitive
2. Set the derivitive to zero and solve for x
3. Test two x points on an interval
4. Determine if the critical numbers are local or absolute max mins

I assumed because this week looks to follow a similar process (at least to an inexperienced individual), I would follow the same steps. Which in the last lesson, I had to solve the quadratic.
 
Hckyplayer8 said:
Are critical points and critical numbers interchangeable?

My previous lesson regarding critical numbers and max/mins followed the process

1. Find the derivitive
2. Set the derivitive to zero and solve for x
3. Test two x points on an interval
4. Determine if the critical numbers are local or absolute max mins

I assumed because this week looks to follow a similar process (at least to an inexperienced individual), I would follow the same steps. Which in the last lesson, I had to solve the quadratic.
Click to expand...
I had to solve the quadratic. What exactly does that mean?????
 
Harry_the_cat said:
You don't need to find when x2-7x+17=0.
Your solution "[7+radical -19] over [2] is correct (except for a +/- sign) and means there are no x-intercepts.
But this is not what the question is asking.

You have correctly solved f '(x) = 0 to get x = 7/2.
This means that there is either a local max, local min, or point of horizontal inflection when x= 7/2.
Your job is to now determine which one of these occurs.

A local minimum occurs when the gradient just to the left of the critical point is pos, and the gradient just to the right of the point is neg.
A local maximum occurs when the gradient just to the left of the critical point is neg, and the gradient just to the right of the point is pos.
If the gradient is pos on both sides or neg on both sides then you have a point of horizontal inflection (PHI).

Consider a value close to but less than x =7/2. Lets say x=3. What is f ' (3)? (you are interested in the sign here rather than the value)
Consider a point close to but more than x = 7/2/ Let's say x = 4. What is f ' (4)? (ditto)
Click to expand...

f ' (3) = -1
f ' (4) = 1
f(3) = 5
f(4) = 5

Thus since there is a negative slope at f(3) and a positive slope at f(4). I went with a local min which graphing the function confirms.

Which means f is decreasing [neg infinity , 7/2) and f is increasing ( 7/2, pos infinity].

I'm still confused about not having to solve the quadratic. For example in this thread I had to solve the quadratic for the zeros. But in this one, I do not? Is it because the other one I was looking for the absolute max and mins?
 
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Jomo said:
I had to solve the quadratic. What exactly does that mean?????
Click to expand...

In a previous problem (that I felt was closely tied to the problem in this thread), I had to solve the given quadratic to find the zeros.
 
Hckyplayer8 said:
I'm still confused about not having to solve the quadratic. For example in this thread I had to solve the quadratic for the zeros …
Click to expand...
Hi. The degree of the given polynomials are not the same in both exercises. As it turned out, the first derivative in this exercise is a linear polynomial, but the first derivative in that other exercise is a quadratic polynomial.

We use the first derivative, when finding critical values. That's why you solved a quadratic equation in the other exercise, but solved a linear equation in this one.

?
 
Hckyplayer8 said:
Are critical points and critical numbers interchangeable?

My previous lesson regarding critical numbers and max/mins followed the process

1. Find the derivitive
2. Set the derivitive to zero and solve for x
3. Test two x points on an interval
4. Determine if the critical numbers are local or absolute max mins

I assumed because this week looks to follow a similar process (at least to an inexperienced individual), I would follow the same steps. Which in the last lesson, I had to solve the quadratic.
Click to expand...
Please ask yourself these questions (to really learn):

What was the "find" of the problem when you followed those four steps?

What was "given" in the problem?

Write (explicitly) "find" and "given" of these problems before deciding on "solution steps".
 
Jomo said:
Are critical points and critical numbers interchangeable? Yes
Click to expand...
I would say, "not exactly- "numbers" are not "points". A "critical number" for function f(x) is a value of x such that f'(x)= 0 or is not defined. A "critical point" is the point (x, f(x)) where x is a "critical number".
 
It depends on whose terminology you are using. Both are used by different people. For example, these use "critical point" to mean a point in the domain (that is, a point on the number line for a function of a real variable):
"Critical number" seems to be less common:
The two are presented interchangeably here:
I don't think I've run across any usages where "critical point" refers to a point on the graph, though that makes sense.
 
Otis said:
Hi. The degree of the given polynomials are not the same in both exercises. As it turned out, the first derivative in this exercise is a linear polynomial, but the first derivative in that other exercise is a quadratic polynomial.

We use the first derivative, when finding critical values. That's why you solved a quadratic equation in the other exercise, but solved a linear equation in this one.

?
Click to expand...

Thank you for posting Otis. I see my error now.
 
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