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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.

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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).

User Michael Erickson
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