Which point on the graph
$\displaystyle y=4x^3+2$
closest to 2,0
I just need a general idea of what to do and Ill do the work
The general idea is illustrated in these threads:
http://www.mathhelpforum.com/math-he...e-problem.html
http://www.mathhelpforum.com/math-he...-x-origin.html
Ok If I use the distance forumla I get
$\displaystyle D=\sqrt{(x-2)^2+(4x^3+2)^2}$
If I simplify the inside which I assume I can I get
$\displaystyle D=\sqrt{16x^6+17x^2+8}$
?
Which when I do the derivative of
$\displaystyle \frac{96x^5+34x}{2\sqrt{16x^6+17x^2+8}}$
then I solve the numerator for 0? which I dont know how I would do that
Almost- you have the square root only around the $\displaystyle (x-7)^2$. It should be around the entire expression. You do that by putting the entire expression in { }. But as TheEmptySet said earlier, to minimize $\displaystyle \sqrt{f(x)}$, it is sufficient to minimize f(x). So find the x and y that minimize $\displaystyle (x- 7)^2+ \left(\frac{12- 2x}{3}- 8\right)^2$.