Question

What is the equation of the tangent line of `f(x)=sqrt(x^2-x+7` at `x=3`?

Answer 1

`2sqrt(13)y = 5x +11`

Explanation:

The slope of of the tangent line is found by differentiating the function. The tangent line is given by `y = mx +c` where `m` is the slope, and the constant can be found from the coordinates of the point on the original curve.

`f'(x) = 1/2(x^2-x+7)^(-1/2)*(2x - 1)`
`f'(x)=(2x-1)/(2sqrt(x^2-x+7))`
At `x=3` `f'(x) = (2*3-1)/(2sqrt(9-3+7)`
`f'(x) = 5/(2sqrt(13)`
At the same point `f(x) = sqrt(9-3+7) = sqrt(13)`

Therefore for the tangent line,
`sqrt(13) = (5/(2sqrt(13)))*3+c`
`c = sqrt(13) - 15/(2*sqrt(13)) = (26-15)/(2*sqrt(13)) =11/(2sqrt(13))`
The tangent line is therefore `y = 5/(2sqrt(13))x +11/(2sqrt(13)`

`2sqrt(13)y = 5x +11`