Question

If the tangent line to y = f(x) at (6, 4) passes through the point (0, 3), find f(6) and f '(6). what's f(6) and f'(6)?

Answer 1

Please see below. Note that you were not asked to find `f(x)`. But I did it for you to give you a better grasp of the concepts. No matter what form the function of `f(x)` has `f(6)=4`.

Explanation:

.

Let's first find the equation of the tangent line from the coordinates of the two given points.

`m=(y_2-y_1)/(x_2-x_1)=(3-4)/(0-6)=(-1)/(-6)=1/6`

`y=mx+b`

We plug in the coordinates of the point of tangency to solve for `b`:

`4=(1/6)(6)+b`

`4=1+b`

`b=4-1=3`

The equation of the tangent line is:

`y=1/6x+3`

To find the slope of the tangent line to the curve, we take the derivative of the function of the curve and evaluate it with the coordinates of the point of tangency.

This means that `f'(6)=m=1/6` which is the slope of the tangent line.

Therefore, one solution would be:

`f'(x)=1/x`

Now, we can take the integral of this function to arrive at the function of the curve:

`f(x)=int(1/x)dx=lnx+C`

We can use the coordinates of the point of tangency to solve for the constant of integration `C`:

`4=ln6+C`

`C=4-ln6`

`f(x)=lnx+4-ln6`

`f(6)=ln6+4-ln6=4`

`f(6)` is the `y` value of the point whose `x`-coordinate is `6` which is the point of tangency.

Another solution would be:

`f'(x)=x^2-6x+1/6`

If we plug in `6` for `x` we get:

`f'(6)=(6)^2-6(6)+1/6=36-36+1/6=1/6` which is the same slope of the tangent line we had before.

Now, if we integrate this function we arrive at the function of our curve which would be:

`f(x)=int(x^2-6x+1/6)dx=1/3x^3-3x^2+1/6x+C`

Now, we can use the coordinates of the point of tangency to solve for `C`:

`4=1/3(6)^3-3(6)^2+1/6(6)+C`

`4=72-108+1+C`

`C=39`

`f(x)=1/3x^3-3x^2+1/6x+39`

In the first solution, we got a natural logarithm function for the curve and in the second solution, we got a cubic curve.

As you can see, there are infinite solutions.