differentiation help (1 Viewer)

[Aerlinn]

I desperately need help with this question, so I hope someone cna help me with it.

I was given a linear graph, which was f'(x) (the derivative function), so I found its equation. Knowing the f'(x) equation, I found the original equation, f(x)= bla bla + C. I was given one piece of information to find what C was, and that was the minimum/ turning point value of x, ie the x coordinate of the turning point of f(x)

How would you approach this question? Can someone outline the steps you'd need to take? I can't figure it out.

Help would be appreciated!
:wave:

 

[menelaus]

what is the actual info you have been given and what is f(x)?

 

[watatank]

Writing an example question would be nice.

 

[pLuvia]

Let's say the linear equation was y=2x+1, this is the derivative of f(x)
f'(x)=2x+1
f(x)=x²+x+C

Now you said that they gave you the coordinates to the minimum turning point, now just sub that x into the f(x) and you can find the constant C.

This was just an example but that's the gist of the method

 

[aussiechica7]

hint: the turning point is when f'(x) = 0.

 

[jyu]

The x-intercept of the linear graph gives the x-coordinate of the turning point. So the given information (I was given one piece of information to find what C was, and that was the minimum/ turning point value of x, ie the x coordinate of the turning point of f(x) ) is redundant. To find the constant C you need another piece of information, e.g. the y-coordinate of the turning point!

 

[jyu]

FInd the x intercept (y=0) to get what C equals

Please explain!

 

[jyu]

Let's say the linear equation was y=2x+1, this is the derivative of f(x)
f'(x)=2x+1
f(x)=x²+x+C

Now you said that they gave you the coordinates to the minimum turning point, now just sub that x into the f(x) and you can find the constant C.

This was just an example but that's the gist of the method

Only the x-coordinate is given!

 

[Aerlinn]

That was what I was thinking, jyu. That's the reason I can't do the question. Is there any way to do it with just the x-coordinate of the tp? =S

 

[-pari-]

can't you still do it?

because you know at a turning point, y' = 0

so if for example, x= 3, then you know that at a point where x = 3, y' = 0.
sub it in...

2x + 1 = 0
when x = 3, 2x + 1 = 0

...hmm...ok well 7=/= 0 ....but isn't there a similar way like this to do it?

 

[nathan71088]

Let's say the linear equation was y=2x+1, this is the derivative of f(x)
f'(x)=2x+1
f(x)=x²+x+C

Now you said that they gave you the coordinates to the minimum turning point, now just sub that x into the f(x) and you can find the constant C.

This was just an example but that's the gist of the method

I believe I have a solution for this:

to find f(x) one would need a "y" and an "x" value to find "C".

Now considering f'(x) = 2x+1
and: f(x) = x²+x+C


at f'(x) = 0 there is a turning point:

2x+1 = 0

x=-1/2

thats one coordinate now:

there is a formula i think it is y= -b/2a. This formula does not need the c considering f(x) = x²+x+C. Therefore

a = 1
and b = 1 ("a" being x^2 coefficient and "b" being x's coefficient)

there fore "y" at the turning point:

y = -(1)
.....2(1) (dots only used to allow for the 2(1) to be shown under the -1)

y = -1/2

Therefore at the turning point x = -1/2 and y = -1/2

Subbing into f(x): (-1/2) = (-1/2)^2 + (-1/2) +C

therefore: 0 = (-1/2)^2 + C

1/4 = C

C = 1/4

therefore: f(x) = x²+x+1/4

 

[Aerlinn]

Yea, you're right. They may have mixed the x and y coordinates up somehow. Isn't there a rule for finding the y-coordinate of the turning point as well? I don't remember what it is though, same as having x= -b/2a for the x coordinate... may help, may not. ..