# Suppose that f(x)= (5x+-1)^1/4 (A) Find an equation for the tangent line to the graph of f(x) at x=2.Tangent line: y = (B) Find all values of x where the tangent line is horizontal, and enter them...

Suppose that f(x)= (5x+-1)^1/4

(A) Find an equation for the tangent line to the graph of f(x) at x=2.

Tangent line: y =

(B) Find all values of x where the tangent line is horizontal, and enter them as a comma separated list (e.g.,* 2,-3,6*). If there are none, enter *none*.

Average of x values =

### 1 Answer | Add Yours

You need to remember what is the form of the equation of tangent line such that:

`y - y_0 = f'(x_0)(x-x_0)`

The problem provides the coordinate `x_0` of tangency poin, hence you need to find `y_0` substituting 2 for x in equation`f(x)= (5x+-1)^1/4` such that:

`f(2)= (10+1)^1/4 =gt f(2) = root(4)11`

You need to differentiate the function with respect to x such that:

`f'(x) = (1/4)*(5x+1)^(1/4 - 1)`

`f'(x) = (1/4)*(5x+1)^(-3/4)`

`f'(x) = (1/4)/(root(4)(5x+1)^(3))`

You need to substitute 2 for x in`f'(x)` such that:

`f'(2) = (1/4)/(root(4)(10+1)^(3))`

`f'(2) = 1/(4root(4)(11)^(3))`

You need to substitute 2 for `x_0` , `root(4)11` for`y_0` and `1/(4root(4)(11)^(3))` for `f'(x_0)` such that:

`y -root(4)11 = 1/(4root(4)(11)^(3))*(x-2)`

`y = 1/(4root(4)(11)^(3))*(x-2) + root(4)11`

**Hence, evaluating the equation of the tangent line to the graph of the given function at `x=2` , yields `y = 1/(4root(4)(11)^(3))*(x-2) + root(4)11.` **