Question

question 14
consider the polynomial function $f(x) = 5x^7 + x^4 + 2x^3 - 8x^8 - 6$.
as $x \to -\infty$, $f(x) \to$ ?
as $x \to \infty$, $f(x) \to$ ?

Explanation:

Step1: Identify Leading Term

The polynomial \( f(x) = 5x^7 + x^4 + 2x^3 - 8x^8 - 6 \) has leading term \( -8x^8 \) (highest degree term). Degree is 8 (even), leading coefficient is -8 (negative).

Step2: Analyze End Behavior (Even Degree)

For a polynomial with even degree:
- If leading coefficient is negative:
- As \( x \to -\infty \), \( x^8 \to \infty \) (since even power), so \( -8x^8 \to -\infty \times \infty = -\infty \)? Wait, no: \( x^8 \) is positive for any \( x \), so \( -8x^8 \) is negative times positive. Wait, no: Wait, when \( x \to -\infty \), \( x^8 = (-x)^8 = x^8 \) (even power), so \( x^8 \to \infty \). Leading term: \( -8x^8 \). So:
- As \( x \to -\infty \): \( x^8 \to \infty \), so \( -8x^8 \to -\infty \)? Wait no, wait: \( x^8 \) is always positive (even exponent), so \( -8x^8 \) is negative times positive, so as \( x \to -\infty \), \( x^8 \) is large positive, so \( -8x^8 \) is large negative? Wait no, wait: Wait, no, let's correct. For even degree \( n \):
- If leading coefficient \( a_n < 0 \):
- As \( x \to \infty \): \( x^n \to \infty \), so \( a_n x^n \to -\infty \) (since \( a_n < 0 \))
- As \( x \to -\infty \): \( x^n = (-x)^n = x^n \) (since \( n \) even), so \( x^n \to \infty \), so \( a_n x^n \to -\infty \) (since \( a_n < 0 \))

Wait, let's re-express the polynomial in standard form (descending degrees): \( f(x) = -8x^8 + 5x^7 + x^4 + 2x^3 - 6 \). Leading term: \( -8x^8 \), degree 8 (even), leading coefficient -8 (negative).

Step3: End Behavior for \( x \to -\infty \)

For even degree, \( x^8 \) is positive when \( x \) is negative (since even power). So \( -8x^8 \): coefficient -8 (negative) times positive \( x^8 \) (which goes to \( \infty \) as \( x \to -\infty \)): so \( -8 \times \infty = -\infty \)? Wait no, wait: \( x \to -\infty \), \( x^8 = (-\infty)^8 = \infty \) (since even), so \( -8x^8 = -8 \times \infty = -\infty \)? Wait, no, wait: Wait, if \( x \) is negative, \( x^8 \) is positive and large, so \( -8x^8 \) is negative and large (since -8 times large positive is large negative). So as \( x \to -\infty \), \( f(x) \approx -8x^8 \to -\infty \)? Wait no, wait, no: Wait, no, wait, let's take an example. Let \( x = -1000 \) (large negative). Then \( x^8 = (-1000)^8 = 1000^8 \) (positive, very large). Then \( -8x^8 = -8 \times 1000^8 \) (very large negative). So as \( x \to -\infty \), \( f(x) \to -\infty \)? Wait no, wait, no, wait: Wait, no, wait, the leading term is \( -8x^8 \), so the end behavior is determined by the leading term. For even degree \( n \):

- If \( a_n > 0 \): as \( x \to \pm\infty \), \( f(x) \to \infty \)
- If \( a_n < 0 \): as \( x \to \pm\infty \), \( f(x) \to -\infty \)

Wait, that's the key. For even degree, the ends (as \( x \to \infty \) and \( x \to -\infty \)) both go in the same direction, determined by the leading coefficient.

So leading term: \( -8x^8 \), degree 8 (even), leading coefficient -8 (negative). So:

- As \( x \to \infty \): \( x^8 \to \infty \), so \( -8x^8 \to -\infty \)
- As \( x \to -\infty \): \( x^8 \to \infty \) (since even), so \( -8x^8 \to -\infty \)

Wait, but let's confirm with the leading term. So:

For \( x \to -\infty \): \( f(x) \approx -8x^8 \to -\infty \) (since \( x^8 \) is large positive, times -8 is large negative)

For \( x \to \infty \): \( f(x) \approx -8x^8 \to -\infty \) (since \( x^8 \) is large positive, times -8 is large negative)

Wait, but let's check with the leading term's degree and coefficient:

- Degree: 8 (even)
- Leading coefficient: -8 (negative)

So end behavior: both ends (as \( x \to \pm\infty \)) go to \( -\infty \)? Wait, no, wait: Wait, no, wait, when degree is even, the left and right ends have the same behavior. If leading coefficient is positive, both go to \( \infty \); if negative, both go to \( -\infty \).

Yes, that's the rule:

- If degree \( n \) is even:
- \( a_n > 0 \): as \( x \to \pm\infty \), \( f(x) \to \infty \)
- \( a_n < 0 \): as \( x \to \pm\infty \), \( f(x) \to -\infty \)

- If degree \( n \) is odd:
- \( a_n > 0 \): as \( x \to \infty \), \( f(x) \to \infty \); as \( x \to -\infty \), \( f(x) \to -\infty \)
- \( a_n < 0 \): as \( x \to \infty \), \( f(x) \to -\infty \); as \( x \to -\infty \), \( f(x) \to \infty \)

So in this case, degree 8 (even), leading coefficient -8 (negative). So:

- As \( x \to -\infty \), \( f(x) \to -\infty \)
- As \( x \to \infty \), \( f(x) \to -\infty \)

Wait, but let's verify with the leading term. Let's take \( x \to \infty \): \( x^8 \to \infty \), so \( -8x^8 \to -\infty \). For \( x \to -\infty \): \( x^8 = (-x)^8 = x^8 \to \infty \), so \( -8x^8 \to -\infty \). So both limits are \( -\infty \).

Answer:

As \( x \to -\infty \), \( f(x) \to \boldsymbol{-\infty} \)
As \( x \to \infty \), \( f(x) \to \boldsymbol{-\infty} \)