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Use the Intermediate Value Theorem to show that the polynomial f(x) =x^3 + x^2 - 4x+ 43 has a real zero between

-5 and -2.

...

Select the correct choice below and fill in the answer boxes to complete your choice.

O A. Because f(x) is a polynomial with f( -5) =
between -5 and -2.

O B. Because f(x) is a polynomial with f( -5) =
between -5 and -2.

OC. Because f(x) is a polynomial with f( -5) =
between -5 and -2.

O D. Because f(x) is a polynomial with f( -5) =
between -5 and -2.

> 0 and f(-2)=

>0, the function has a real zero

> 0 and f(-2)=

<0, the function has a real zero

<0 and f(-2)=

<0, the function has a real zero

< 0 and f(-2)=

>0, the function has a real zero

1 answer

To apply the Intermediate Value Theorem (IVT) to the polynomial \( f(x) = x^3 + x^2 - 4x + 43 \), we first need to evaluate \( f(x) \) at the endpoints \( x = -5 \) and \( x = -2 \).

  1. Calculate \( f(-5) \): \[ f(-5) = (-5)^3 + (-5)^2 - 4(-5) + 43 = -125 + 25 + 20 + 43 = -37 \]

  2. Calculate \( f(-2) \): \[ f(-2) = (-2)^3 + (-2)^2 - 4(-2) + 43 = -8 + 4 + 8 + 43 = 47 \]

Now we have:

  • \( f(-5) = -37 \) (which is \( < 0 \))
  • \( f(-2) = 47 \) (which is \( > 0 \))

Since \( f(-5) < 0 \) and \( f(-2) > 0 \), by the Intermediate Value Theorem, there must be at least one real zero of \( f(x) \) in the interval \((-5, -2)\).

Thus, we can choose the correct option:

Correct choice: A. Because \( f(x) \) is a polynomial with \( f(-5) < 0 \) and \( f(-2) > 0 \), the function has a real zero between -5 and -2.