Finding Max and Mins Part II

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Hckyplayer8

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Determine the absolute max and min values of f(x)=x3+x2-8x+5 on the interval [-3,4] and the x values where they occur.

Step one is find the derivative.

f(x)=x3+x2-8x+5
f'(x)=3x2+2x-8

We are dealing with a quadratic, so I figured we are going that route to solve for the zeros.

My quadratic calculations came out to x=4/3 for the positive equation and x=-2 for the negative.

The next step is to substitute the bounded x values on the interval and the x values derived from the quadratic, back into the equation.

f(-3) = 11
f(4) = 53
f(-2) = 17
f(4/3) = -41/27

Thus the max of the function is at point [4,53] and the min is at point [4/3,-41/27]
 
The correct procedure with respect to a closed interval is to: (1) find all points where the first derivative is zero on the open interval, (2) determine whether those points define a local minimum or a local minimum on the open interval (by the second derivative test if that works), (3) compute the value of the function at each local minimum and maximum, (4) compute the values of the function at the two end points, and (5) specify the location and value of the greatest of the maxima and the least of the minima.
 
Hckyplayer8 said:
So that is the correct process and answer?
Click to expand...
To me that’s sounds quite good. Just be careful not all roots to the first derivative have to be a local max or min. Always remember y=x^3…y’=3x^2… x=0 is not a max nor min, if you take the second derivative you see 0 is its root too which means its a Inflection Point.
 
JeffM said:
The correct procedure with respect to a closed interval is to: (1) find all points where the first derivative is zero on the open interval, (2) determine whether those points define a local minimum or a local minimum on the open interval (by the second derivative test if that works), (3) compute the value of the function at each local minimum and maximum, (4) compute the values of the function at the two end points, and (5) specify the location and value of the greatest of the maxima and the least of the minima.
Click to expand...

My teacher did not go over a second derivative test. Can you please specify?

Going over your post, I think my problem looks correct.
 
Well then the only thing you need to know now is when you find a root for the first derivative it might not be a max or a minimum. Imagine a local maximum point. We go up until we reach that point and then we go down, right?
Ok it means right before we reach the point we have a positive slope (imagine a -x^2 graph for example) then we reach the slope of 0 at the point ( slope,derivative =0) ,after that we go down and go to negative slope

So just check the sign of the derivative (negative or positive) around and near the root to see whether they are maximum minimum or inflection point
This could be helpful ;)
 
Hckyplayer8,
Jagulba is talking about classifying stationary points as max, min or points of horizontal inflection. This is important stuff.
BUT, to find absolute max and min in a given domain, as in your question, you need only compare the y-values of points at the extreme ends of the domain and where f '(x) =0.
 
Harry_the_cat said:
Hckyplayer8,
Jagulba is talking about classifying stationary points as max, min or points of horizontal inflection. This is important stuff.
BUT, to find absolute max and min in a given domain, as in your question, you need only compare the y-values of points at the extreme ends of the domain and where f '(x) =0.
Click to expand...
Thanks ;) You define my points quite well :)))
 
Hckyplayer8 said:
My teacher did not go over a second derivative test. Can you please specify?

Going over your post, I think my problem looks correct.
Click to expand...
If f(x) is differentiable on (a, b) and there is a point c such that a < c < b and f(c) is a local minimum or maximum, f'(c), meaning the value of the first derivative at x = c, will equal zero. HOWEVER, not every value of x where f'(x) = 0 indicates a value of x where f(x) is a local maximum or minimum. If you want to find out whether any people in a room are mothers, you need to question only the women in the room, but that does not mean that every woman in the room is a mother. It is the difference between a necessary condition and a sufficient condition. With me to here?

Furthermore, even if f(c) is a local minimum or maximum, it is true that f'(c) = 0, but that does not tell you whether f(c) is a maximum or a minimum, which is usually a very important point. So finding where f'(x) = 0 is just one step in finding the location and value of a local maximum or minimum.

Although there are various ways to finish the process, one way (that is available only if f'(x) is differentiable at x = c) is to find the value of f''(c), where f''(x) is the second derivative of f(x), which is defined as the derivative of the first derivative of f(x). If f'(c) = 0 and f''(c) > 0, then f(x) has a local minimum at x = c. If f'(c) = 0 and f''(c) < 0, then f(x) has a local maximum at x = c. If, however, f'(c) = 0 = f''(c), the second derivative test fails, and more than one situation may be the case. What happens then can get a bit complex, but one very common situation is that c is a point of inflection for f(x).
 
JeffM said:
If f(x) is differentiable on (a, b) and there is a point c such that a < c < b and f(c) is a local minimum or maximum, f'(c), meaning the value of the first derivative at x = c, will equal zero. HOWEVER, not every value of x where f'(x) = 0 indicates a value of x where f(x) is a local maximum or minimum. If you want to find out whether any people in a room are mothers, you need to question only the women in the room, but that does not mean that every woman in the room is a mother. It is the difference between a necessary condition and a sufficient condition. With me to here?

Furthermore, even if f(c) is a local minimum or maximum, it is true that f'(c) = 0, but that does not tell you whether f(c) is a maximum or a minimum, which is usually a very important point. So finding where f'(x) = 0 is just one step in finding the location and value of a local maximum or minimum.

Although there are various ways to finish the process, one way (that is available only if f'(x) is differentiable at x = c) is to find the value of f''(c), where f''(x) is the second derivative of f(x), which is defined as the derivative of the first derivative of f(x). If f'(c) = 0 and f''(c) > 0, then f(x) has a local minimum at x = c. If f'(c) = 0 and f''(c) < 0, then f(x) has a local maximum at x = c. If, however, f'(c) = 0 = f''(c), the second derivative test fails, and more than one situation may be the case. What happens then can get a bit complex, but one very common situation is that c is a point of inflection for f(x).
Click to expand...

Thank you for the explanation.
 
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