Business Calculus Question

Question

Let

𝑓(𝑥)= (x-5)/(3x+4)

Find an equation for the tangent line to the graph of f at 𝑥=11

Answer 1

Hi Ashley,

Step 1 is to find the y-coordinate of your point.

f(11) = (11 - 5) / (3*11 + 4) = 6/37

This is a garbage fraction and I'm not sure why the question writers made this question more difficult than they had to.

Anyway.

To find the slope of the tangent line at that point, you need to find f'(x) and plug in 11. You'll need to use the quotient rule.

f(x) = (x - 5) / (3x + 4)

Let's say:

g(x) = x - 5

h(x) = 3x + 4

So

g'(x) = 1

h'(x) = 3

We can rewrite f(x) as:

f(x) = g(x) / h(x)

By the Quotient Rule:

f'(x) = [h(x)*g'(x) - g(x)*h'(x)] / (h(x))^2

So,

f'(x) = [(3x+4)(1) - (x-5)(3)] / (3x+4)^2

= [3x+4 - 3x + 15] / (3x+4)^2

= 19 / (3x+4)^2

Plug in 11 for x to find the slope of the tangent line at 11.

f'(11) = 19 / (3*11+4)^2

= 19 / 37^2

= 19 / 1369

So the slope of your tangent line is 19/1369. See comment above about question writers.

So we have a point P(11, 6/37), and a slope m= 19/1369. We can use point-slope form to find the equation of the line.

y - 6/37 = 19/1369 * (x - 11)

I don't think we should bother trying to clean up that mess. The question writers should be compelled to stare at that and think about what they did.

Hope this helps!