To determine the coordinates of the point on the graph of f(x) = sqrt(2x+1) where the tangent line is perpendicular to the line 3x+y+4=0, we first need to find the slope of the tangent line.
The line 3x+y+4=0 is in the form of y = mx + b, where m is the slope. So we rewrite the equation as y = -3x - 4.
For a line to be perpendicular to another line, the slopes must be negative reciprocals of each other. Therefore, the slope of the tangent line (m1) is the negative reciprocal of the slope of the given line (m2).
The slope of the given line (m2) is -3, so the slope of the tangent line (m1) is 1/3.
Next, we find the derivative of f(x) = sqrt(2x+1) to get the slope of the tangent line at any given point on the graph.
f(x) = sqrt(2x+1)
f'(x) = (1/2√(2x+1))(2)
= 1/√(2x+1)
Now, we equate the derivative to 1/3 and solve for x:
1/√(2x+1) = 1/3
Cross-multiply:
√(2x+1) = 3
Square both sides:
2x+1 = 9
2x = 9-1
2x = 8
x = 4
So, the x-coordinate of the point on the graph of f(x) = sqrt(2x+1) is 4.
To find the y-coordinate, we substitute x = 4 into the equation f(x) = sqrt(2x+1):
f(4) = sqrt(2(4)+1)
= sqrt(8+1)
= sqrt(9)
= 3
So, the y-coordinate of the point on the graph is 3.
Therefore, the coordinates of the point where the tangent line is perpendicular to the line 3x+y+4=0 is (4, 3).