Question

How do you find the points where the graph of the function `(3x^3) - (14x^2) -(34x) + 347` has horizontal tangents and what is the equation?

Answer 1

Points are `x=4.045` and `x=-0.934` and tangents are `y=178.955` and `y=364.0999`

Explanation:

Slope of a tangent is given by `(df(x))/(dx)` and tangent is horizontal when slope is `0`.

In otherwords, one will find horizontal tangents, where `(df)/(dx)=0`

As `f(x)=3x^3-14x^2-34x+347`, we have

`(df)/(dx)=9x^2-28x-34`

and `9x^2-28x-34=0`

i.e. `x=(28+-sqrt(28^2-4*9*(-34)))/18`

= `(28+-sqrt(784+1224))/18`

= `(28+-44.81)/18`

= `72.81/18` or `-16.81/18`

= `4.045` or `-0.934`

Further when `x=4.045`, `f(x)=178.955`

and when `x=-0.934`, `f(x)=364.099`

Hence tangents are `y=178.955` and `y=364.0999`