# What is the point where tangent line at the curve y = e^x(x-1) is parallel to the acute bisectrix?

### 1 Answer | Add Yours

You need to remember that the tangent line to the curve, at a point, is given by the derivative of the function at that point.

You also need to remember that `f'(x_0) = m` , where m represents the slope of the tangent line to the curve.

The problem requests the points where the tangent line is parallel to the acute bisectrix, hence, you need to use the equation that relates the slopes of two parallel lines, such that:

`m_1 = m_2`

`m_1` represents the slope of tangent line

`m_2` represents the slope of acute bisectrix

You need to write the equation of acute bisectrix, such that:

`y = x `

You need to identify the slope `m_2,` such that:

`m_2 = 1` (coefficient of variable x)

You need to evaluate `m_1` , hence, you need to evaluate the first derivative of the given function `y = e^x(x - 1)` , using the product rule, such that:

`y' = (e^x)'(x - 1) + e^x(x - 1)'`

`y' = e^x(x - 1) + e^x => y' = e^x(x - 1 + 1) => y' = e^x*x=> m_1 = e^x*x`

You need to set the following equation, such that:

`m_1 = m_2 => e^x*x = 1 => e^x = 1/x`

You need to solve for x the transcendental equation, using the graphical method, hence, you need to sketch the graphs of the functions `y = e^x` and `y = 1/x` , such that:

You need to notice that the black curve representing the function `y = e^x` intersects the red curve representing the function `y = 1/x` , at a point `x in (0,1)` .

**Hence, evaluating the x coordinate of the point where the tangent line to the curve `y = e^x(x - 1)` is parallel to the acute bisectrix, yields `x in (0,1)` .**