find a function y=f(x) whose second derviative is y''=12x-2 at each point (x,y) on its graph and y=-x+5 is tangent to the graph at the point corresponding to x=1

Question

zepdrix · Answer

Hmm this is a neat problem :)

So ummm.... y=-x+5 is tangent to the curve at x=1.
The `slope` of this tangent line gives us the value of the first derivative at x=1.

$\Large\rm y'(1)=-1$

Understand how I got that information?

anonymous · Answer

yes a little thanks

zepdrix · Answer

You can integrate / anti-differentiate and that data will allow you to find the unknown that shows up.

zepdrix · Answer

Integrating a second time to our y function brings another constant popping up.
So we need to find another piece of data we can use to figure out that constant.

zepdrix · Answer

Recall that a tangent line `touches` the curve.
So the tangent line and the function y will both share a coordinate pair.

So plug x=1 into the tangent line,
$\Large\rm y=-(1)+5$
This will give us our coordinate pair that we can use to find the other constant term.

anonymous · Answer

so the would the final result ne f(x)= 2x^3+x^2-5x+4

zepdrix · Answer

Hmm I came up with something a little different.

anonymous · Answer

what did you come up with

zepdrix · Answer

$$\Large\rm y''=12x-2$$Anti-differentiating gives us:$$\Large\rm y'=6x^2-2x+c$$Plugging in our data,$$\Large\rm y'(1)=-1 \qquad\to\qquad -1=6-2+c\qquad\to\qquad c=-5$$

zepdrix · Answer

So our first derivative is,$$\Large\rm y'=6x^2-2x-5$$

zepdrix · Answer

Anti-differentiating again gives us,$$\Large\rm y=2x^3-x^2-5x+c_1$$

zepdrix · Answer

The tangent line and our curve both pass through the point (1,4).

$$\Large\rm 4=2-1-5+c_1\qquad\to\qquad c_1=8$$Something like that I think?

zepdrix · Answer

$$\Large\rm y=2x^3-x^2-5x+8$$

anonymous · Answer

yeah I had did something wrong with but I get it now thank you so much :)

zepdrix · Answer

cool \c:/