On the graph f(x)=x^2/3x find the point that is closest to point (3,0).?

Answer 1

y = f(x)=x²/3x = x/3
y = x/3

The closest point will be on the line thru the point (3,0) and perpendicular to the line y = x/3.

The slope of the given line is 1/3. So the slope of the perpendicular line is -1/(1/3) = -3.

The equation of the perpendicular line is:

y - 0 = -3(x - 3)
y = -3x + 9

Solving for the intersection of the two lines we have:

y = x/3
y = -3x + 9

0 = (10/3)x - 9
(10/3)x = 9
x = (3/10)(9) = 27/10

y = x/3 = (27/10)(1/3) = 9/10

So the point closest to (3,0) is (27/10,9/10).

Answer 2

Your f(x) looks a little ambiguous. Remember the importance of brackets when asking these questions, i.e. do you mean

(x^(2/3))x, or do you mean (x^2)/(3x)? I'm going to assume you mean (x^2)/(3x).

f(x) = (x^2)/3x

This actually reduces; you can cancel out an x, to get

f(x) = x/3

To find the point closest to (3,0), we plug in the the values (3,0) and (x,y) into the distance formula.

d = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 )
d = sqrt ( (x - 3)^2 + (y - 0)^2 )

But we know that y = x/3, so

d = sqrt ( (x - 3)^2 + (x/3)^2 )
d = sqrt ( (x - 3)^2 + (1/9)(x^2) )

So this will be our distance function, d(x).

d(x) = sqrt ( (x - 3)^2 + (1/9)(x^2) )

Let's square this function, since, in order to minimize the distance, we're going to calculate the derivative d'(x) and make it 0 anyway.

[d(x)]^2 = (x - 3)^2 + (1/9)(x^2)

Taking the derivative of both sides (implicitly, with respect to x), we get

2[d(x)][d'(x)] = 2(x - 3) + (2/9)x

Making d'(x) = 0, we get

0 = 2(x - 3) + (2/9)x
Multiplying both sides by 9,

0 = 18(x - 3) + 2x
0 = 18x - 54 + 2x
0 = 20x - 54
54 = 20x, therefore x = 54/20 = 27/10

Now, we found our x-coordinate closest to (3,0). We find our y-coordinate by plugging in x = 27/10 in the function
f(x) = x/3
f(27/10) = 27/30 = 9/10

Therefore, the point closest to (3,0) on the graph f(x) = (x^2)/(3x) is (27/10, 9/10)

Answer 3

Pythagorean therm a^2+b^2=c^2
using f(x) and the point (0,3)
let a=difference in x=delta x= always| x-3 |

let b=difference in y =delta y = |f(x)-0|=|(x^2/3x)-0|=|x/3|

(x-3)^2+(x/3)^2= c^2
foil the powers

x^2-6x+9 +( x^2/9)=1 2/9x^2 -6x+9= distance^2=c^2
deritive of the distance function |11(x^-2)/9-6x+9| =|-22(x^-3)-6|=0 when -22/x^3=6 so -22=6x^3 and -22/6=x^3 or -11/3= x^3 or x=cube root of -11/3 or x = 2/3

so I think (2/3, 4)

I am not sure how to find the deritive of negative powers so double check this part

Answer 4

It may help to simplify your equation for f(x) = x/3 It would help to see your textbook to know what technique you are expected to use. I am thinking that you would need to write the distance equation so that you would have the distance from point (3, 0) for all values of x. Then you would need to differentiate the resulting equation and find the zero to give the value of x for the minimum distance.

Answer 5

y=(3)^2/(3*3)
y=1

therefore, the point 3,1 is closest.