For the function f(x) = x/(3x+1) find the equation of the line which is tangent to the curve f(x) at x=2. Solve analytically (an exact answer is required)
Please show steps.
Thanks.
y–y0=f'(x0)*(x–x0)
f'=1/(3x+1)^2
x0=2
y0=2/(3*2+1)=2/(6+1)=2/7
f'(x0)=1/(3*2+1)^2=1/(6+1)^2=1/7^2=1/49
y–y0=f'(x0)*(x–x0)
y - 2/7 = 1/49*(x-2)
y=(1/49)*x - 2/49 + 2/7
y=(1/49)*x - 2/49 + 14/49
y=(1/49)*x - 12/49
y=(1/49)*(x-12)
If You don't know how find derivation of x/(3x+1) in google type: calc101
When you open page calc101com click option derivatives.In rectangle type:
x/(3x+1)
and click option DO IT
You will see solution step by step
To find the equation of the line that is tangent to the curve f(x) at x=2, we need to first find the slope of the curve at that point. The slope of the tangent line can be obtained by taking the derivative of f(x) with respect to x, and then evaluating it at x=2.
Step 1: Find the derivative of f(x)
The derivative of f(x) can be found using the quotient rule. Given f(x) = x / (3x+1), we differentiate the numerator and denominator separately and apply the quotient rule:
f'(x) = ( (3x+1) * 1 - x * 3 ) / (3x+1)^2
= (3x + 1 - 3x) / (3x+1)^2
= 1 / (3x+1)^2
Step 2: Evaluate the derivative at x=2
To find the slope of the tangent line at x=2, substitute x = 2 into the derivative equation:
f'(2) = 1 / (3(2)+1)^2
= 1 / (6+1)^2
= 1 / 7^2
= 1 / 49
Step 3: Use the point-slope form to write the equation of the tangent line
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line. Since the tangent line passes through the point (2, f(2)), we substitute x1 = 2 and y1 = f(2) = 2 / (3*2+1) = 2/7 into the equation:
y - (2/7) = (1/49)(x - 2)
This is the equation of the line that is tangent to the curve f(x) at x=2.