Question

How do I solve this?

Find all points on the graph 𝑦 = x^3*e^2x where the tangent line to the graph is horizontal.

Answer 1

`(0,0), (-3/2,-27/8e^3)`. See below.

Explanation:

The tangent line to the graph of the function is horizontal wherever the first derivative of the function is equal to `0.`

Graphically interpreted, the derivative tells us the slope, or rate of change, of the function at a given point. Furthermore, we draw tangent lines to the curve to give a more concise visual representation of the slope. A derivative of zero implies we have a rate of change of zero and therefore a horizontal tangent line.

Find `y'`, set it equal to `0,` and solve for `x:`

`y'=2e^(2x)x^3+3x^2e^(2x)` (Product Rule and Chain Rule)

`2e^(2x)x^3+3x^2e^(2x)=0`
Take out a common factor of `x^2e^(2x):`

`x^2e^(2x)(2x+3)=0`

We now have two equations to solve for `x:`

`x^2e^(2x)=0`

`2x+3=0`

For `x^2e^(2x)=0:`

`e^(2x)` can never equal `0`; this is one of the fundamental properties of the exponential function. Since we don't have to worry about it being equal to `0` at any point, we can divide out `e^(2x)` and be left with `x^2`

`x^2=0`

Taking the square root of both sides yields

`x=0`

So, the tangent line is horizontal at `x=0.` To find the `y`-value of this point, simply plug in `0` into `y:`

`y(0)=0^3*e^(2*0)=0`

The tangent line is horizontal at `(0,0)`

For `2x+3=0`

`2x=-3`

`x=-3/2`

The tangent line is horizontal at `x=-3/2`. Plug this value into `y` to obtain the full coordinate pair:

`y(-3/2)=(-3/2)^3e^(2*3/2)=-27/8e^3`

The tangent line is horizontal at `(-3/2,-27/8e^3)`

In decimal, to three significant figures, this is `(-1.50, -0.168)`. We can confirm our two points visually by checking a graph: