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help quick calculus (1 Viewer)

astroman

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the function has one turning point. Find its coordinates and show that it is a maximum turning point.

i'm getting confused on 'one turning point' coz i've only been calculating stationary points until this question.:confused:
 

SquareHeartsAdrita

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the turning point it the vertex of the parabola aka the stationary point. you can use calculus to get that point by solving f'(x) = 0 or alternatively the coordinates will be (-b/2a, f(-b/2a) )

to show it is a max; show f''(-b/2a) < 0
 
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<a href="http://www.codecogs.com/eqnedit.php?latex=f(x)=&space;7-4x-x^{2}\\&space;f'(x)=-4-2x&space;\\&space;$Stationary/Turning&space;Points&space;occur&space;when&space;first&space;derivative&space;=&space;0&space;\\&space;-4-2x&space;=&space;0&space;\\&space;\therefore&space;x=-2&space;\\&space;\text{since&space;only&space;one&space;x&space;value,&space;hence&space;only&space;one&space;turning&space;point}&space;\\&space;f''(x)=&space;-2&space;\\&space;f''(x)<&space;0&space;\therefore&space;\text{Maximum&space;turning&space;point}&space;\\&space;\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x)=&space;7-4x-x^{2}\\&space;f'(x)=-4-2x&space;\\&space;$Stationary/Turning&space;Points&space;occur&space;when&space;first&space;derivative&space;=&space;0&space;\\&space;-4-2x&space;=&space;0&space;\\&space;\therefore&space;x=-2&space;\\&space;\text{since&space;only&space;one&space;x&space;value,&space;hence&space;only&space;one&space;turning&space;point}&space;\\&space;f''(x)=&space;-2&space;\\&space;f''(x)<&space;0&space;\therefore&space;\text{Maximum&space;turning&space;point}&space;\\&space;\\" title="f(x)= 7-4x-x^{2}\\ f'(x)=-4-2x \\ $Stationary/Turning Points occur when first derivative = 0 \\ -4-2x = 0 \\ \therefore x=-2 \\ \text{since only one x value, hence only one turning point} \\ f''(x)= -2 \\ f''(x)< 0 \therefore \text{Maximum turning point} \\ \\" /></a>

When testing whether a point is max or min, sub the x value into second derivative, if <0 then you have a max turning point, if >0 then you have a minimum turning point
 

zhertec

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Turning points are stationary points, there are two types of stationary points: turning and inflection. Turning being obviously change in gradient, while inflection being a change in concavity (I think lol).

To find the total number of stationary points dash the first equation and let that equal 0, hence finding the x values for the stationary points. This will be the x value for the coordinate of the max point. Sub into original equation to acquire the y coordinate.

Now to test for max or min, there are two ways: One is a sure fire method, but a tad more tedious, while the other is easier but sometimes does not work in some cases (when the double dashed equals 0 I think).

The easier method:
Dash the first equation twice in order to obtain a numerical value in which >0 is equal to a min point and <0 is a max point.

The second better method:
Since you know the x value of the turning points (in which its gradient equals to zero), test if there is a change in gradient. So basically x-e (to discover if the gradient left of the stationary point is positive or negative) and sub that into the dashed equation to discover the gradient and then x+e (to determine the gradient, at right of the stationary point). Where e= any small numerical value you can choose (generally 0.001 or something, just as long as it's small enough to not pass another stationary point).

Btw put this in a table form as it is a lot easier to see. Then for every y' value discovered by subbing in the x-e/x+e put it in the table and determine where it is positive or negative. Giving a positive gradient a "/" so you know it is rising. And a negative gradient "\" so you know it is decreasing.
Eventually you see a pattern

\ -- / = min turning point

/ -- \ = max turning point

--=the gradient of the stationary point, which is 0.
 
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answer is (-2,11) and f'(x)>0
plz explain some1
Yeah i forgot lol, when you get the x value, sub it into the original y equation to get the y coordinate. Hence (-2,11). Im not sure what you mean by f'(x)>0, what do you want to know?
 

zhertec

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If this pic sucks, it's only this is my first time uploading a pic lol

P.S. can anyone see the attachment? lol if not anyone got a quick guide :3

P.S.S. phew did it.

Mafs.JPG
 
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