Questions: If f''(x) = x(x+1)(x-2)^2, then the graph of f has inflection points when x= ? (A) -1 only (B) -1 and 0 only (C) -1 and 2 only (D) -1, 0, 2 only

Transcript text: If $f^{\prime \prime}(x)=x(x+1)(x-2)^{2}$, then the graph of $f$ has inflection points when $x=$ ? (A) -1 only (B) -1 and 0 only (C) -1 and 2 only (D) -1, 0, 2 only

Solution

Solution Steps

To find the inflection points of the function $ f $ given its second derivative $ f''(x) = x(x+1)(x-2)^2 $, we need to determine where $ f''(x) $ changes sign. This occurs at the roots of $ f''(x) $, which are the points where $ f''(x) = 0 $. We then check the intervals around these roots to see where the sign changes.

Solution Approach

Find the roots of $ f''(x) = x(x+1)(x-2)^2 $.
Check the sign of $ f''(x) $ around these roots to determine where it changes sign.
The points where the sign changes are the inflection points.

Step 1: Find the Roots of $ f''(x) $

To find the inflection points, we first need to determine where the second derivative $ f''(x) = x(x + 1)(x - 2)^2 $ is equal to zero. Setting $ f''(x) = 0 $ gives us the equation:

\[ x(x + 1)(x - 2)^2 = 0 \]

The roots of this equation are found by solving for $ x $:

\[ x = 0, \quad x = -1, \quad x = 2 \]

Step 2: Check for Sign Changes

Next, we check the sign of $ f''(x) $ around the roots to identify where the function changes concavity. We evaluate $ f''(x) $ at points slightly to the left and right of each root:

For $ x = -1 $:
- $ f''(-1 - 0.1) $ is positive.
- $ f''(-1 + 0.1) $ is negative.
This indicates a sign change at $ x = -1 $.
For $ x = 0 $:
- $ f''(0 - 0.1) $ is negative.
- $ f''(0 + 0.1) $ is positive.
This indicates a sign change at $ x = 0 $.
For $ x = 2 $:
- $ f''(2 - 0.1) $ is positive.
- $ f''(2 + 0.1) $ is positive.
There is no sign change at $ x = 2 $.