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
I do not seem to get the answer... I find the first derivative of f(x0 and equate it to the value of the slope from the tangent line. the answer says it should be (4,3) but i do not even get the x-value correct.
my derivatiave is (1/2)(2x+1)^(-1/2)(2)
= 1/√(2x+1) , that is the slope of the tangent.
slope of 3x+y+4 = -3
so the slope of the perpendicular is 1/3
then 1/√(2x+1) = 1/3
square both sides and crossmultiply
2x+1 = 9
x = 4
sub x=4 into original to finish
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).
To determine the coordinates of the point on the graph of f(x) = √(2x + 1) where the tangent line is perpendicular to the line 3x + y + 4 = 0, you need to follow these steps:
1. Start by finding the first derivative of f(x). The derivative of √(2x + 1) is (1/2) * (2x + 1)^(-1/2).
2. Next, you need to find the slope of the line 3x + y + 4 = 0. To do this, rearrange the equation in the form y = mx + b, where m is the slope. In this case, the given line can be rewritten as y = -3x - 4, so the slope is -3.
3. Since the tangent line is perpendicular to the given line, its slope will be the negative reciprocal of -3, which is 1/3.
4. Set the derivative equal to 1/3 to find the x-coordinate of the point where the tangent line is perpendicular to the given line: (1/2) * (2x + 1)^(-1/2) = 1/3.
5. Solve this equation to find the x-coordinate. Multiply both sides by 2 and square both sides to eliminate the square root:
(2x + 1)^(-1/2) = 2/3
[(2x + 1)^(-1/2)]^2 = (2/3)^2
2x + 1 = 4/9
2x = 4/9 - 1
2x = -5/9
x = -5/18
6. Now that you have the x-coordinate, substitute it back into the original equation f(x) = √(2x + 1) to find the corresponding y-coordinate:
f(-5/18) = √(2(-5/18) + 1) = √(-5/9 + 9/9) = √(4/9) = 2/3
So the coordinates of the point on the graph of f(x) where the tangent line is perpendicular to the line 3x + y + 4 = 0 are (-5/18, 2/3).
It seems like there might be an error in the answer you mentioned (4,3). Please double-check your calculations and make sure you followed the steps correctly.