Question

At what point on the given curve is the tangent line parallel to the line 5x - y = 5? y = 4 + 2ex − 5x

Answer 1

`(\ln5, 14-5\ln5)`

Explanation:

Equation of given curve:

`y=4+2e^x-5x`

`\frac{d}{dx}y=\frac{d}{dx}(4+2e^x-5x)`

`\frac{dy}{dx}=2e^x-5`

The above derivative `\frac{dy}{dx}` shows the slope of tangent line at any point to the given curve.

It is given that the tangent to the curve is parallel to the line: `5x-y=5` or `y=5x-5` hence the slope of tangent will be equal to the slope of line i.e. `5`

hence, the slope of tangent `\frac{dy}{dx}` must be equal to `5` as follows

`\frac{dy}{dx}=5`

`2e^x-5=5`

`e^x=5`

`x=\ln5`

Substituting `x=\ln5` in equation of curve to get y-coordinate of point of contact as follows

`y=4+2e^{\ln5}-5\ln5`

`=4+2\cdot 5-5\ln 5`

`=14-5\ln5`

hence, the point of contact on the given curve is `(\ln5, 14-5\ln5)`