Math

Question Find the values of xx where the function y=6(x1)4(x2)2(x+1)y=6(x-1)^{4}(x-2)^{2}(x+1) changes sign.

Studdy Solution

STEP 1

Assumptions
1. The function given is y=6(x1)4(x2)2(x+1) y = 6(x-1)^{4}(x-2)^{2}(x+1) .
2. We need to determine at which value of x x the function changes sign.

STEP 2

Identify the zeros of the function, which are the values of x x that make the function equal to zero. These are found by setting each factor in the function to zero.

STEP 3

Set the first factor equal to zero and solve for x x .
(x1)4=0 (x-1)^{4} = 0

STEP 4

Since the factor is raised to the fourth power, the zero occurs with even multiplicity, meaning the graph of the function touches the x x -axis at this point but does not cross it.
x=1 x = 1

STEP 5

Set the second factor equal to zero and solve for x x .
(x2)2=0 (x-2)^{2} = 0

STEP 6

Again, since the factor is raised to the second power, the zero occurs with even multiplicity, meaning the graph of the function touches the x x -axis at this point but does not cross it.
x=2 x = 2

STEP 7

Set the third factor equal to zero and solve for x x .
(x+1)=0 (x+1) = 0

STEP 8

Since the factor is not raised to any power greater than one, the zero occurs with odd multiplicity, meaning the graph of the function crosses the x x -axis at this point.
x=1 x = -1

STEP 9

Now, we analyze the behavior of the function at each zero. For zeros with even multiplicity (such as x=1 x = 1 and x=2 x = 2 ), the function does not change sign because the graph touches the x x -axis and turns back.

STEP 10

For the zero with odd multiplicity (such as x=1 x = -1 ), the function does change sign because the graph crosses the x x -axis.

STEP 11

Therefore, the function y=6(x1)4(x2)2(x+1) y = 6(x-1)^{4}(x-2)^{2}(x+1) changes sign at x=1 x = -1 .
The function changes sign at x=1 x = -1 .

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