Question

in the prob. find pt on graph of f(x)=x^2-1 closest to (-4,0) i plugged pts. and eqn. into distance formula and then took d/dx to get 4x^3-2x+8, do i need to solve for 0? how?

Answer 1

i'm only getting non-real results

Answer 2

the distance formula is\[\sqrt{\left( x+4 \right)^{2}+\left( x ^{2} -1\right)^{2}}\]
and would require the chain rule.\[_{2}^{1}\left( x ^{4}-x ^{2} +8x+17 \right)^{-1/2}\left( 4x ^{3}-2x+8 \right)\]
Hope that helps the rest is ugly.

Answer 3

I found both a real and imaginary answer to your question using dx/dy as opposed to the conventional dy/dx. Since there is a real answer by simple observation then my real answer is my solution.

Answer 4

Maveric's first response was spot on. Taking what would become the simplified numerator from the derivative shown you can get the one real solution while ignoring the two complex roots. Unfortunately, I can't factor:\[2x ^{3}-x+4=0\], but graphing it on my calculator, I get a zero crossing at x=-1.392 corresponding to y=0.937 at a distance of 2.771. Maveric's second response has some errors and results in a distance of 6.0266. I tried desperately to find a manual method to factor, but even Wolfram says that there is no simplified factorization.

Answer 5

Here is another approach....
use the point-slope form of a line and the slope of the normal line to the graph of the quadratic as the value for m. You now have an equation of the "line" loosely defined of course, that intersects the quadratic at the point of interest. Then set them equal and solve. The solution yields the same result as presented above \[2x^3-x+4=0\].
Using this method, you can enter the functions into a graphing calculator and find the points of intersection. I hope this helps.