Question

A differentiable function f satisfies f(3) = 5, f(9) = 7, f′(3) = 11, and f′(9) = 13.
Find an equation for the tangent line to the curve
y = f(x2) at the point (x,y) = (3, 7).

Answer 1

equation of any line is:
y=bx+c
where b is the slope, and c deteminns the position of the line in the x-y plane.
The tangent line to the curve f(x) at some point )x_0,y_0) has it's slope, which is f'(x_0).
so: y=f'(x_0)x+c
Now you need o find the posiion of this line to find c.

Answer 2

For that, just notice that it has to pass through the point (x_0,y_0).
Putting in the values:
slope is: \(2xf'(x^2)\). This has to be calclated at point (3,7). Which gives \(6f'(9)\).
From the information you have on the function, this is equal 78.
So the equation of the line is taking form, y=78x+c
To find c, notice that it is a result of this:
\(y-y_0=b(x-x_0)\) this is equation of the line that passes through the point (x_0,y_0)
from here \(y=bx+(y_0-bx_0)\)
so you see that c is the second term of the equality on the right side.
I think tou can do it from here