F(x)= (x-7)/(9x-1) find f’(x) and find the equation of the line tangent to the graph of f at x=2

Answer 1

To find f'(x), we need to differentiate the function f(x). Let's find the derivative step by step:

f(x) = (x - 7) / (9x - 1)

Using the quotient rule, we have:

f'(x) = (9x - 1)((1) - (x - 7)(9)) / (9x - 1)^2

Next, we simplify:

f'(x) = (9x - 1 - 9x^2 + 63x - 63) / (9x - 1)^2

f'(x) = (-9x^2 + 72x - 64) / (9x - 1)^2

To find the equation of the line tangent to the graph of f at x = 2, we need to find the slope of the tangent line at x = 2, and then use the point-slope equation of a line to write the equation.

To find the slope at x = 2, we substitute x = 2 into f'(x):

f'(2) = (-9(2)^2 + 72(2) - 64) / (9(2) - 1)^2

f'(2) = (-36 + 144 - 64) / (18 - 1)^2

f'(2) = 44 / 289

Now, let's use the point-slope equation of the line to write the equation:

y - y1 = m(x - x1)

We have the point (2, f(2)) = (2, f(2)) and the slope m = f'(2).

Finding f(2):

f(x) = (x - 7) / (9x - 1)

f(2) = (2 - 7) / (9(2) - 1)

f(2) = -5 / 17

Therefore, the point (2, f(2)) = (2, -5/17).

Now we can write the equation of the line tangent to the graph of f at x = 2:

y - (-5/17) = (44 / 289)(x - 2)

y + 5/17 = (44 / 289)(x - 2)

y + 5/17 = (44/289)x - (88/289)

y = (44/289)x - (88/289) - 5/17

y = (44/289)x - (88/289) - (85/289)

y = (44/289)x - (88/289) - (85/289)

y = (44/289)x - (173/289)

So, the equation of the line tangent to the graph of f at x = 2 is y = (44/289)x - (173/289).