An equation of the line tangent to the graph of (5x - 1)/(3x + 1) at the point where x = 1 is? Any ideas?
sure, I have an idea. Find the derivative to get the slope of the tangent line! Then use the point-slope form of the line's equation:
y = (5x-1)/(3x+1)
y(1) = 4/4 = 1
y' = 8/(3x+1)^2
y'(1) = 8/16 = 1/2
tangent line is thus
(y-1) = 1/2 (x-1)
To find the equation of the line tangent to the graph of a function at a specific point, you need to find the derivative of the function and evaluate it at that point.
Let's start by finding the derivative of the function (5x - 1)/(3x + 1) using the quotient rule.
The quotient rule states that if we have a function f(x) = g(x)/h(x), then its derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2
In this case, g(x) = 5x - 1 and h(x) = 3x + 1.
Now, let's find the derivatives of g(x) and h(x).
g'(x) = 5 (since the derivative of x is 1)
h'(x) = 3 (since the derivative of x is 1)
Now, let's apply the quotient rule formula to find f'(x):
f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2
= (5 * (3x + 1) - (5x - 1) * 3)/((3x + 1)^2)
= (15x + 5 - 15x + 3)/(9x^2 + 6x + 1)
= 8/(9x^2 + 6x + 1)
Next, we need to evaluate f'(x) at x = 1 to find the slope of the line tangent at that point.
f'(1) = 8/(9(1)^2 + 6(1) + 1)
= 8/(9 + 6 + 1)
= 8/16
= 1/2
So, the slope of the tangent line at x = 1 is 1/2.
Now, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is the point of tangency and m is the slope.
In this case, the point of tangency is (1, (5(1) - 1)/(3(1) + 1)) = (1, 4/4) = (1, 1).
Plugging the values into the point-slope form, we have:
y - 1 = (1/2)(x - 1)
Simplifying, we get:
y - 1 = (1/2)x - 1/2
Finally, rearranging the equation to the standard form:
y = (1/2)x + 1/2
Therefore, the equation of the line tangent to the graph of (5x - 1)/(3x + 1) at the point where x = 1 is y = (1/2)x + 1/2.
To find the equation of the line tangent to a graph at a given point, you need to determine the slope of the graph at that point and use the point-slope form of a line. Here's how you can find the equation of the tangent line:
1. Determine the derivative of the function: Let's first find the derivative of the given function f(x) = (5x - 1)/(3x + 1). The derivative gives us the slope of the tangent line at any given point on the graph.
To find the derivative, we can use the quotient rule or take advantage of the fact that the derivative of a constant times a function is equal to the constant times the derivative of the function. Using the latter method, we have:
f'(x) = [ (d/dx)(5x - 1) * (3x + 1) - (5x - 1) * (d/dx)(3x + 1) ] / (3x + 1)^2
The derivative of (5x - 1) is simply 5, and the derivative of (3x + 1) is 3. Plugging these values into the equation, we get:
f'(x) = (5 * (3x + 1) - 3 * (5x - 1)) / (3x + 1)^2
2. Substitute the x-coordinate of the given point into the derivative: Since we want to find the tangent line at x = 1, substitute x = 1 into the derivative equation:
f'(1) = (5 * (3 * 1 + 1) - 3 * (5 * 1 - 1)) / (3 * 1 + 1)^2
= (5 * 4 - 3 * 4) / 4^2
= (20 - 12) / 16
= 8 / 16
= 1/2
So, the slope of the tangent line at the point x = 1 is 1/2.
3. Use the point-slope form to find the equation of the tangent line: We have the slope (m = 1/2) and the point (x = 1, f(1)), which we can find by substituting x = 1 into the original function:
f(1) = (5 * 1 - 1) / (3 * 1 + 1)
= 4 / 4
= 1
So, the point on the graph is (1, 1).
Now we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope:
y - 1 = (1/2)(x - 1)
Expanding and rearranging the equation:
y - 1 = 1/2x - 1/2
y = 1/2x + 1/2
Therefore, the equation of the tangent line to the graph of (5x - 1)/(3x + 1) at the point x = 1 is y = 1/2x + 1/2.