Find the point on the curve which is closest to the point (7,0)

(4.5, 5.1)

Step-by-step explanation:

The distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

Then if we want to find the point on the curve y = √(5*x + 3) that is closest to the point (7, 0) we need to minimize the distance:

D = √( (x - 7)^2 + (y - 0)^2)

D = √( (x - 7)^2 + (√(5*x + 3))^2)

Because we know that D is positive, minimizing D is the same than minimizing D^2

Then we can minimize:

D^2 = (x - 7)^2 + (5*x + 3)

D^2 = x^2 - 14*x + 21 + 5*x + 3

D^2 = x^2 - 9*x + 24

This is a quadratic equation with a positive leading coefficient, then the minimum of this function is at the vertex.

To find the vertex, we need to find the zero of the first derivative, this is:

(D^2)' = 2*x - 9

We need to solve:

0 = 2*x - 9

9 = 2*x

9/2 = x

4.5 = x

To find the correspondent y-value, we need to evaluate the curve in x = 4.5

y = √(5*4.5 + 3)  = 5.1

Then the point is (4.5, 5.1)

This means that the point on the curve y = √(5*x + 3) which is closest to the point (7, 0) is the point (4.5, 5.1)

idk

step-by-step explanation: