Questions: Consider the following polynomial function. f(x)=(x+2)^4(x-1)^3(x-3) Step 3 of 3 : Find the zero(s) at which f'' flattens out. Express the zero(s) as ordered pair(s).

Consider the following polynomial function.
f(x)=(x+2)^4(x-1)^3(x-3)

Step 3 of 3 : Find the zero(s) at which f'' flattens out. Express the zero(s) as ordered pair(s).

Transcript text: Consider the following polynomial function. \[ f(x)=(x+2)^{4}(x-1)^{3}(x-3) \] Step 3 of 3 : Find the zero(s) at which $f^{\prime \prime}$ flattens out". Express the zero(s) as ordered pair(s).

Solution

Solution Steps

Step 1: Find the second derivative $f''(x)$

The second derivative of the given polynomial is: $f''(x) = 4 x \left(14 x^{5} + 126 x^{4} + 180 x^{3} + 30 x^{2} - 33 x - 9\right)$

Step 2: Solve $f''(x) = 0$

Step 3: Verify inflection points

An inflection point is confirmed if the sign of $f''(x)$ changes before and after the value of $x$. The confirmed inflection points are:

$x = 0$
$x = -7.28$
$x = -1.33$
$x = -0.5$
$x = -0.32$
$x = 0.42$

Step 4: Express the inflection points as ordered pairs

The inflection points as ordered pairs $(x, f(x))$ are:

$(0, 0)$
$(-7.28, -1694021.15)$
$(-1.33, 8.91)$
$(-0.5, 0.16)$
$(-0.32, 0.08)$
$(0.42, -0.55)$

Final Answer:

The inflection point at $x = 0$ is $(0, 0)$. The inflection point at $x = -7.28$ is $(-7.28, -1694021.15)$. The inflection point at $x = -1.33$ is $(-1.33, 8.91)$. The inflection point at $x = -0.5$ is $(-0.5, 0.16)$. The inflection point at $x = -0.32$ is $(-0.32, 0.08)$. The inflection point at $x = 0.42$ is $(0.42, -0.55)$.

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