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minimum and maximum value help (1 Viewer)

hungry

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Doing 2-unit bridging course and have no idea how to do this, I kind of got half the answer but even then I probably did it wrong. My working might look weird or be wrong because I used several different sources which did this a different way. And also because I don't know how to math at all.
^ = squared

QUESTION:

Find the minimum/maximum value of:

f(x) = 4x - x^

What I did so far:

f(x) = 4x - x^
f(x) = 4 - 2x
f(0) = 4 - 2x

0 = 4
x = 4

I don't know how to do the rest, and I know what I have isn't right. I've mixed stuff up and don't know what I'm doing. But I know 4 is part of the answer, since it's in the answers page. So I'm getting somewhere, I guess.

Can someone please help me do this question, and please be patient and also be detailed in your answer - you're helping someone who failed year 10 maths.
 

photastic

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Test nature using the second derivative. Sub the x value and you will determine its nature. Once subbed in, if the value is >0 it's a minimum, whereas if the value is <0 it's a maximum.

You can also determine nature using the table and subbing values between the x value.
 
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BlueGas

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1. Find the stationary points, this is when you differentiate the equation and make f(x) = 0.

2. Use the second derivative to find out whether x=4 is a maximum or a minimum, the second derivative is -2 and this is less than zero, so it is a maximum because the second derivative is less than zero

3. If there's anything you don't understand, just say.
 

hungry

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1. Find the stationary points, this is when you differentiate the equation and make f(x) = 0.

2. Use the second derivative to find out whether x=4 is a maximum or a minimum, the second derivative is -2 and this is less than zero, so it is a maximum because the second derivative is less than zero

3. If there's anything you don't understand, just say.
Yeah I don't really understand anything, mainly because I get confused with the math terms as I am unfamiliar with them. Do you think you could (or anyone reading this) solve it and show the steps with brief explanations? So I can try to do it with a different question?
 

BlueGas

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Yeah I don't really understand anything, mainly because I get confused with the math terms as I am unfamiliar with them. Do you think you could (or anyone reading this) solve it and show the steps with brief explanations? So I can try to do it with a different question?
What math terms are hard for you, just so I can get an understanding and know what to explain to you. The steps I mentioned in my above post can be used for any sort of question that asks you to find maximum/minimum.
 

hungry

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What math terms are hard for you, just so I can get an understanding and know what to explain to you. The steps I mentioned in my above post can be used for any sort of question that asks you to find maximum/minimum.
I don't understand how to 'differentiate' an equation and also how to get the second derivative
 

photastic

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I don't understand how to 'differentiate' an equation and also how to get the second derivative
Differentiate the first derivative to get to the second derivative. Differentiating is when you minus the power and multiple the original power.
 

BlueGas

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I don't understand how to 'differentiate' an equation and also how to get the second derivative
Very simple, let's use a few examples.

1. 3x^2

For this example, you multiply the power to whatever number is infront of the x. So 3x2 = 6, after you multiply the power to the number infront of the x, you minus the power by 1, so 2-1 = 1. So the final answer is 6x.

2. x^3.

For this example, it's not different to the first, maybe because there's no number infront of the x that might confuse you, but basically when you just have the x by itself there is a 1 infront of the x. So for this question, you do you 3x1 = 3, and then you minus the power by 1, so your final answer 3x^2.

3. 4x^3 - 6x.

For this example, you start differentiating from the right of the equation, so let's first start by differentiating 4x^3. You first do 4x3 which equals 12 and then you minus one from the power so now you have 12x^2. Then you differentiate -6x. Now let me mention something, when you just have a number and x with no power, you basically just take away the x, and when you differentiate -6x, you're just left with -6, get it now?
 

BlueGas

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To find the second derivative, you just differentiate the equation you differentiated, if that makes sense. So using an example above, we'll use 3x^2.

First derivative = 6x.

To find the second derivative, you differentiate the first derivative (6x).

So when you differentiate 6x you get 6, and basically 6 is your second derivative.
 

hungry

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I keep getting 0, was my original working out that got me 4 correct?
 

BlueGas

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I keep getting 0, was my original working out that got me 4 correct?
Ahh I didn't pay attention to your working out, the actual answer is x=2. What you did wrong is that you subbed in x=0 into the first derivative, you're not meant to do that.

What I meant above when I said make "f(x)=0" is that you should do 0=4 - 2x. Since f(x) = 4 - 2x, you just make f(x) = 0 to find what x equals.

To find what x equals, you have to get it by itself, in this case, you bring the four to the other side so since it's +4 you minus for so 0 - 4 = -4.

Now -4 = -2x.

Between -2 and x is a multiply sign, and to find x, you have to get it by itself, so it bring the -2 over to the other side, you divide by -2 so -4/-2 = 2.

Therefore x=2.

Hopefully you'll be able to understand that.
 

hungry

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Ahh I didn't pay attention to your working out, the actual answer is x=2. What you did wrong is that you subbed in x=0 into the first derivative, you're not meant to do that.

What I meant above when I said make "f(x)=0" is that you should do 0=4 - 2x. Since f(x) = 4 - 2x, you just make f(x) = 0 to find what x equals.

To find what x equals, you have to get it by itself, in this case, you bring the four to the other side so since it's +4 you minus for so 0 - 4 = -4.

Now -4 = -2x.

Between -2 and x is a multiply sign, and to find x, you have to get it by itself, so it bring the -2 over to the other side, you divide by -2 so -4/-2 = 2.

Therefore x=2.

Hopefully you'll be able to understand that.
Thank you that helped a lot, and I think I'm starting to get the hang of it. But I seem to be doing just one thing wrong, I do get one of the values correct, but then the next one is wrong. The following is a NEW example, not the one I used previously. Can you please tell me what I'm doing wrong here:

f(x) = x^2 - 4x
f(x) = 2x - 4x

0 = 2x - 4
0 + 4 = 2x
4 = 2x
4 + 2x

4/2 = 2/1 = 2 (this value seems to be correct)

f(x) = (2 * 1) - 1^2 = 1 (this is the part I think I'm doing wrong).
 

BlueGas

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Thank you that helped a lot, and I think I'm starting to get the hang of it. But I seem to be doing just one thing wrong, I do get one of the values correct, but then the next one is wrong. The following is a NEW example, not the one I used previously. Can you please tell me what I'm doing wrong here:

f(x) = x^2 - 4x
f(x) = 2x - 4x

0 = 2x - 4
0 + 4 = 2x
4 = 2x
4 + 2x

4/2 = 2/1 = 2 (this value seems to be correct)

f(x) = (2 * 1) - 1^2 = 1 (this is the part I think I'm doing wrong).
Yes that's the part you are doing wrong. After you have found what x equals, you find the second derivative.

So basically differentiate 2x - 4 which equals 2.

Since you have no x in the second derivative, it makes working out whether it's a maximum/minimum slight easier for you.

Once you find your second derivative, you see if it's greater or less than zero. If it's greater than zero, it's a minimum, if it's less than zero, it's a maximum. So for this question, at x=2 it is a minimum.

HERE'S A TIP FOR THE FUTURE:

Sometimes (depending on the question), you might or might not have x in your second derivative.

1. If there's no x, (like the question above), you just see if it's greater or less than zero

2. If there's x, (like for example you get a second derivative of 2x - 1), you sub the x you got when f(x)=0 into the second derivative. so let's say x=2 you do this: 2 x 2 -1, and that's about it.
 

funnytomato

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What I did so far:

f(x) = 4x - x^2
f'(x) = 4 - 2x

f(0) = 4 - 2x

0 = 4
x = 4

I don't know how to do the rest, and I know what I have isn't right. I've mixed stuff up and don't know what I'm doing. But I know 4 is part of the answer, since it's in the answers page. So I'm getting somewhere, I guess.
This might be a lengthy and "time-wasting" explanation, but I couldn't see any "short cut" way. It may take while to understand something, but once you do, it will save you much more time.

It seems that you know the algebraic rule of differentiation (i.e. bring the index down and reduce its index by 1 ),
but not quite sure what this derivative thing ( the expression you get is denoted as ,called the derivative of ) actually means.

It is normal to get confused about notations [ such as , etc. ] at first, don't worry too much about these.
Btw,I don't think the bridging course is a good idea , as they introduce too many new definitions and notations in such a short period of time.

Anyways, let's look your question:

Find the minimum/maximum value of

The derivative as you've found is , and geometrically this expression represents the gradient(or slope, which is
equal to , see the diagram below) of the tangent line(the line that's "touching" the curve)



And one thing we notice is that it changes with different values of x.
Say at , then the y value is . The gradient is . So this might give you an idea of what f(x) and f'(x) actually mean.
Say at a different point where , the gradient would be . (you could sketch it and find out the rise and run)

So when asked to find the maximum/minimum, we're looking for that point where the gradient is 0. (again, diagram might help, try and draw the tangent line at the vertex and you could see that the rise being 0, hence gradient = 0 )

Then we're looking for where (with a "flat" gradient), so you're solving , from which you get . (agreeing with the diagram).

And it says find the max/min, they mean the max/min of y value(i.e f(x)).
Hence at , the max/min is . (again, in agreement with the diagram).

To figure out whether it's maximum or minimum. You can either find the 2nd derivative as photastic mentioned.

Or you can find out the gradients points to left and right of that possible max or min.
To the left, say we pick , we have .
To the right, if we choose , we have .
So we can determine whether it's max or min depending on how the gradients evolve around that point.

 
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BlueGas

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I think I might have also confused you, when I said "f(x)=0", I actually mean f'(x)=0, f'(x) is the first derivative, and f''(x) is the second derivative.
 

hungry

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Yes that's the part you are doing wrong. After you have found what x equals, you find the second derivative.

So basically differentiate 2x - 4 which equals 2.

Since you have no x in the second derivative, it makes working out whether it's a maximum/minimum slight easier for you.

Once you find your second derivative, you see if it's greater or less than zero. If it's greater than zero, it's a minimum, if it's less than zero, it's a maximum. So for this question, at x=2 it is a minimum.

HERE'S A TIP FOR THE FUTURE:

Sometimes (depending on the question), you might or might not have x in your second derivative.

1. If there's no x, (like the question above), you just see if it's greater or less than zero

2. If there's x, (like for example you get a second derivative of 2x - 1), you sub the x you got when f(x)=0 into the second derivative. so let's say x=2 you do this: 2 x 2 -1, and that's about it.
Ahhh thank you so much, I think I've got it now. Thanks both Bluegas and Photastic for helping me, I'm really slow at maths and I've literally been trying to work this out all last night and this morning but nothing made sense until now. I'm so happy right now. :D
 

hungry

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This might be a lengthy and "time-wasting" explanation, but I couldn't see any "short cut" way. It may take while to understand something, but once you do, it will save you much more time.

It seems that you know the algebraic rule of differentiation (i.e. bring the index down and reduce its index by 1 ),
but not quite sure what this derivative thing ( the expression you get is denoted as f'(x),called the derivative of f(x) ) actually means.

It is normal to be confused about notations [ such as f(x) , f'(x) etc. ] at first, don't worry too much about these.
Btw,I don't think the bridging course is a good idea , as they introduce too many new definitions and notations in such a short period of time.

Anyways, let's look your question:

Find the minimum/maximum value of f(x) = 4x - x^2.

The derivative as you've found is f'(x)=4-2x, and geometrically this expression represents the gradient(or slope, which is
, see the diagram below) of the tangent line(the line that's "touching" the curve)



And one thing we notice is that it changes with different values of x.
Say at x=1, then the y value f(1)=4x1-1^2=3. The gradient f'(1)=4-2x1=2. So this might give you an idea of what f(x) and f'(x) actually mean.
Say at a different point where x=0, the gradient would be f(0)=4-2x0=4. (you could sketch it and verify)

So when asked to find the maximum/minimum, we're looking for that point where the gradient is 0. (again, diagram might help, try and draw the tangent line at the vertex and you could see that the rise being 0, hence gradient = 0 )

Then we're looking for where f'(x)=4-2x=0 (with a "flat" gradient), so you're solving , from which you get . (agreeing with the diagram).

And it says find the max/min, they mean the max/min of y value(i.e f(x)).
Hence at , the max/min is . (again, in agreement with the diagram).

To figure out whether it's maximum or minimum. You can either find the 2nd derivative as photastic mentioned.

Or you can find out the gradients points to left and right of that possible max or min.
To the left, say we pick , we have .
To the right, if we choose , we have .
So we can determine whether it's max or min depending on how the gradients evolve around that point.

Oops I didn't refresh the page when I posted my other reply. But thank you for the detailed response, it is very helpful and especially for someone of my level it's very clear and makes sense. :)
 

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