Question

pt. on graph of f(x)=x^2-1 closest to (-4,0)?
i'm getting non-real #'s when solving d/dx for 0

Answer 1

can I see the equation you set up?

Answer 2

d=√(x-4)^2+(x^2-1-0)^2
actually i think i had (x+4)

Answer 3

now i get (√5/2,3/2)?

Answer 4

Well let me work it out. You want a distance equation , which is sqrt( (x - x0)^2 + (y - y0)^2 ) . x0 and y0 are your points. 'x' is just 'x', and 'y' is your curve function.

Answer 5

$$\sqrt{(-4+x)^2+\left(-1+x^2\right)^2}$$

Answer 6

Then you take the derivative of that and get $$\frac{-4-x+2 x^3}{\sqrt{17+x \left(-8-x+x^3\right)}}$$

Answer 7

Set that to 0 and solve and I get one real answer, 1.392

Answer 8

i think its √(x-(-4))^2+((x^2-1)-0) and distance will be least when result inside radical is least, so derive x^4-x^2-8x+17

Answer 9

hmm, from looking at the plot, that doesn't look right; I did something wrong

Answer 10

Ah, I got a minus sign reversed. Ahd you're right, we can ignore the radial when minimizing. So we want to minimize
$$Sqrt[(4 + x)^2 + (-1 + x^2)^2]$$
which is the same as minimizing
$$(4+x)^2+\left(-1+x^2\right)^2$$

Answer 11

Deriving, expanding and collecting I get we want to now set $$8-2 x+4 x^3$$ to 0

Answer 12

Which is -1.392