Math  /  Algebra

QuestionConsider the following polynomial function. f(x)=x4+6x3+x242x56f(x)=x^{4}+6 x^{3}+x^{2}-42 x-56
Step 4 of 4: Find the zero(s) at which ff "flattens out". Express the zero(s) as ordered pair(s).
Answer
Select the number of zero(s) at which ff "flattens out".
Selecting an option will display any text boxes needed to complete your answer. none 1 2 3 4

Studdy Solution

STEP 1

What is this asking? Find the points where the graph of this function flattens out, which means finding the points where the slope of the tangent line is zero! Watch out! Remember, a "flattening out" point isn't necessarily a maximum or minimum.
It could be a point of inflection, like in x3x^3 at x=0x=0.

STEP 2

1. Find the derivative
2. Find critical points
3. Determine which critical points correspond to "flattening out"

STEP 3

Alright, so we're given this funky function f(x)=x4+6x3+x242x56f(x) = x^4 + 6x^3 + x^2 - 42x - 56.
To find where it "flattens out," we need its derivative!
The derivative tells us the instantaneous rate of change, which is the slope of the tangent line at any point.

STEP 4

Using the power rule, the derivative of xnx^n is nxn1nx^{n-1}.
So, the derivative of our function, which we'll call f(x)f'(x), is: f(x)=4x3+18x2+2x42f'(x) = 4x^3 + 18x^2 + 2x - 42 Boom! We just found the derivative!

STEP 5

Critical points are where the derivative is zero or undefined.
Our derivative is a polynomial, so it's defined everywhere.
We just need to find where it's equal to zero.
Setting f(x)f'(x) equal to zero gives us: 4x3+18x2+2x42=04x^3 + 18x^2 + 2x - 42 = 0

STEP 6

We can **divide the entire equation by 2** to simplify it a bit: 2x3+9x2+x21=02x^3 + 9x^2 + x - 21 = 0

STEP 7

Now, we need to find the **roots of this cubic equation**.
By trying some small integer values, we find that x=1x=1 is a root, since 2(1)3+9(1)2+121=2+9+121=02(1)^3 + 9(1)^2 + 1 - 21 = 2 + 9 + 1 - 21 = 0.
Yay!

STEP 8

Since x=1x=1 is a root, we know (x1)(x-1) is a factor.
We can perform polynomial long division or synthetic division to find the other factor.
Dividing 2x3+9x2+x212x^3 + 9x^2 + x - 21 by (x1)(x-1) gives us 2x2+11x+212x^2 + 11x + 21.
So, our factored derivative is: (x1)(2x2+11x+21)=0(x-1)(2x^2 + 11x + 21) = 0

STEP 9

Now, we need to find the roots of 2x2+11x+21=02x^2 + 11x + 21 = 0.
We can use the **quadratic formula**: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=2a=2, b=11b=11, and c=21c=21.
Plugging those values in, we get: x=11±112422122=11±1211684=11±474x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 2 \cdot 21}}{2 \cdot 2} = \frac{-11 \pm \sqrt{121 - 168}}{4} = \frac{-11 \pm \sqrt{-47}}{4} Since we have a negative number under the square root, these roots are complex numbers, and we're only looking for real-valued critical points.
So, our only real critical point is x=1x=1.

STEP 10

We found one critical point: x=1x=1.
This is where the function *might* flatten out.
To be sure, we need to find the corresponding y-value.
Plugging x=1x=1 back into the original function f(x)f(x), we get: f(1)=(1)4+6(1)3+(1)242(1)56=1+6+14256=90f(1) = (1)^4 + 6(1)^3 + (1)^2 - 42(1) - 56 = 1 + 6 + 1 - 42 - 56 = -90

STEP 11

So, the point where the function flattens out is (1,90)(1, -90).

STEP 12

There is **one** zero at which ff "flattens out," and it is (1,90)(1, -90).

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