Question

let be defined for all x by f(x) = x ^ 3 + 3/2* x ^ 2 - 6x + 10 find the stationary points of fand determine the intervals where fincre / 5 / 3 find the inflection point for f.

Answer 1

The function
`\sf f(x)` is defined as:

`\sf f(x) = x^3 + (3)/(2)x^2 - 6x + 10`

To find the stationary points of
`\sf f`, we need to find the values of
`\sf x` where the derivative of
`\sf f(x)` is equal to zero.

First, let's find the derivative of
`\sf f(x)`:

`\sf f'(x) = 3x^2 + 3x - 6`

To find the stationary points, we set
`\sf f'(x) = 0` and solve for
`\sf x`:

`\sf 3x^2 + 3x - 6 = 0`

We can factor the quadratic equation as follows:

`\sf 3(x^2 + x - 2) = 0`

Now, we solve for
`\sf x` by factoring further:

`\sf 3(x + 2)(x - 1) = 0`

This gives us two solutions:
`\sf x = -2` and
`\sf x = 1`.

So, the stationary points of
`\sf f(x)` are
`\sf x = -2` and
`\sf x = 1`.

To determine the intervals where
`\sf f(x)` is increasing, we need to analyze the sign of the derivative
`\sf f'(x)` in different intervals. We can use the values of
`\sf x = -2`,
`\sf 1`, and any other value between them.

For
`\sf x < -2`, we choose
`\sf x = -3` as a test point:

`\sf f'(-3) = 3(-3)^2 + 3(-3) - 6 = 12 > 0`

For
`\sf -2 < x < 1`, we choose
`\sf x = 0` as a test point:

`\sf f'(0) = 3(0)^2 + 3(0) - 6 = -6 < 0`

For
`\sf x > 1`, we choose
`\sf x = 2` as a test point:

`\sf f'(2) = 3(2)^2 + 3(2) - 6 = 18 > 0`

From the above analysis, we can conclude that
`\sf f(x)` is increasing in the intervals
`\sf (-\infty, -2)` and
`\sf (1, \infty)`.

To find the inflection point of
`\sf f`, we need to determine where the concavity changes. This occurs when the second derivative of
`\sf f(x)` changes sign.

The second derivative of
`\sf f(x)` is:

`\sf f''(x) = 6x + 3`

To find the inflection point, we set
`\sf f''(x) = 0` and solve for
`\sf x`:

`\sf 6x + 3 = 0`

`\sf 6x = -3`

`\sf x = -(1)/(2)`

Therefore, the inflection point of
`\sf f(x)` is
`\sf x = -(1)/(2)`.