Question

If `y= x^3 + 6x^2 + 7x -2cosx`, what are the points of inflection of the graph f (x)?

Answer 1

Inflection Point `(-1.8941, -0.5273)`

Explanation:

.

`y=x^3+6x^2+7x-2cosx`

Here is the graph of it:

As you can see, it has an absolute maxima, an absolute minima, and a point of inflection that lies between the extremas.

Its roots are, `x=-4.45418, -1.58116, and 0.230748` which are the points where the graph crosses the `x`-axis.

To find the `x` values of the maximum and minimum points, we take the derivative of the function and set it equal to zero, and solve for its roots:

`y'=3x^2+12x+7+2sinx=0`

Here is the graph of this derivative function:

Its roots are `x=-3.2601 and -0.575411` which are the `x` coordinates of the maximum and minimum points of your original function.

The point of inflection lies between these two extremas and can be found by taking the second derivative of your original function and setting it equal to zero, and solving for its root:

`y''=6x+12+2cosx=0`

`3x+6+cosx=0`

Here is the graph of this function:

It has one root, `x=-1.8941` which is the `x` coordinate of your inflection point. To find its `y` coordinate you have
to plug this into your original function and calculate `y`:

`y=(-1.8941)^3+6(-1.8941)^2+7(-1.8941)-2cos(-1.8941)=-6.7953+21.5256-13.2587-1.9989=-0.5273`

Inflection Point `(-1.8941, -0.5273)`

The above solution was arrived at by using graphing utilities. Without using graphing utilities, it would take some cumbersome algebra in the complex domain to solve for the above `x` values.