For what values of x does the graph of f have a horizontal tangent? f(x) = e^x(cos(x))

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We first take the derivative of f. 

We apply the Product Rule for differentiation here:

`d(uv) = vdu+ udv`

Let set `u = e^x`

So, `du = e^xdx`

` `

Let set `v= cosx`

So, `v = -sinx`

Applying the formula:````

`f ' = e^x*cosx - sinx * e^x`

`f' = e^xcosx -...

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We first take the derivative of f. 

We apply the Product Rule for differentiation here:

`d(uv) = vdu+ udv`

Let set `u = e^x`

So, `du = e^xdx`

` `

 

Let set `v= cosx`

So, `v = -sinx`

Applying the formula:````

`f ' = e^x*cosx - sinx * e^x`

`f' = e^xcosx - sinxe^x`

We can factor out e^x on right side. 

`f' = e^x(cosx - sinx)`

We know that the slope for the horizontal tangent is zero. 

So, we equate f ' to zero. 

`e^x(cosx - sinx) = 0`

We equate each factor to zero. 

`e^x = 0 ; cosx - sinx = 0`

The e^x will never be equal to zero, the range of it is always greater than zero. Hence, we only consider the second equation. 

`cosx - sinx = 0`

We square both sides.

`(cosx - sinx)^2 = 0^2`

`cos^2x - 2sinxcosx + sin^2x = 0`

Using the identity: `cos^2x + sin^2x = 1`

We will have,
`1 - 2sinxcosx = 0` 

USe the double angle for sine: `2sinxcosx = sin(2x)`

```1 - sin(2x) = 0`

Add sin(2x) on each sides.

`sin(2x) = 1`

Take the arcsine of both sides. Note that if the x is in the interval [0, 2pi), the 2x will be in the interval [0, 4pi).

`2x = pi/2` ;`2x = (5pi)/2`

Divide each equation by 2 both sides.

`x = {pi/4, (5pi)/4}`

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