find the points or points on the graph of y= squareroot(3x+40) with -10<=x<=10 that is/are closest to the origin. Havent done calc in a while and I cant find an example in my book to illustrate how to do this problem. thanks for your help! :)
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d = √(x^2 + y^2) = √(x^2 + 3x + 40)
D = d^2 = x^2 + 3x + 40
dD/dx = 2x + 3 = 0
x = -3/2
y = √(3(-3/2) + 40) = √(-9/2 + 40) = √(71/2)
(-3/2, √(71/2)) is the point closest to the origin.
D = d^2 = x^2 + 3x + 40
dD/dx = 2x + 3 = 0
x = -3/2
y = √(3(-3/2) + 40) = √(-9/2 + 40) = √(71/2)
(-3/2, √(71/2)) is the point closest to the origin.