Question

example 1
determine the consecutive integer values of x between which each real zero of each function is
located by using a table. then sketch the graph.

1. ( f(x) = x^2 + 3x - 1 )
2. ( f(x) = -x^3 + 2x^2 - 4 )
3. ( f(x) = x^3 + 4x^2 - 5x + 5 )
4. ( f(x) = -x^4 - x^3 + 4 )

Answer:

Let's solve these problems one by one. We'll use the Intermediate Value Theorem, which states that if a function \( f(x) \) is continuous on an interval \([a, b]\), and \( f(a) \) and \( f(b) \) have opposite signs, then there is at least one real zero of the function between \( a \) and \( b \).

Problem 1: \( f(x) = x^2 + 3x - 1 \)

This is a quadratic function (degree 2), so it has at most 2 real zeros. Let's evaluate \( f(x) \) at integer values of \( x \):

• For \( x = -4 \): \( f(-4) = (-4)^2 + 3(-4) - 1 = 16 - 12 - 1 = 3 \)
• For \( x = -3 \): \( f(-3) = (-3)^2 + 3(-3) - 1 = 9 - 9 - 1 = -1 \)
• For \( x = -2 \): \( f(-2) = (-2)^2 + 3(-2) - 1 = 4 - 6 - 1 = -3 \)
• For \( x = -1 \): \( f(-1) = (-1)^2 + 3(-1) - 1 = 1 - 3 - 1 = -3 \)
• For \( x = 0 \): \( f(0) = 0^2 + 3(0) - 1 = -1 \)
• For \( x = 1 \): \( f(1) = 1^2 + 3(1) - 1 = 1 + 3 - 1 = 3 \)

Now, we look for sign changes:

• Between \( x = -4 \) ( \( f(-4) = 3 \) ) and \( x = -3 \) ( \( f(-3) = -1 \) ): Sign changes from positive to negative, so there's a zero between \( -4 \) and \( -3 \).
• Between \( x = 0 \) ( \( f(0) = -1 \) ) and \( x = 1 \) ( \( f(1) = 3 \) ): Sign changes from negative to positive, so there's a zero between \( 0 \) and \( 1 \).

Problem 2: \( f(x) = -x^3 + 2x^2 - 4 \)

This is a cubic function (degree 3), so it has at least 1 real zero (and at most 3). Let's evaluate \( f(x) \) at integer values of \( x \):

• For \( x = -2 \): \( f(-2) = -(-2)^3 + 2(-2)^2 - 4 = 8 + 8 - 4 = 12 \)
• For \( x = -1 \): \( f(-1) = -(-1)^3 + 2(-1)^2 - 4 = 1 + 2 - 4 = -1 \)
• For \( x = 0 \): \( f(0) = -0^3 + 2(0)^2 - 4 = -4 \)
• For \( x = 1 \): \( f(1) = -1^3 + 2(1)^2 - 4 = -1 + 2 - 4 = -3 \)
• For \( x = 2 \): \( f(2) = -2^3 + 2(2)^2 - 4 = -8 + 8 - 4 = -4 \)
• For \( x = 3 \): \( f(3) = -3^3 + 2(3)^2 - 4 = -27 + 18 - 4 = -13 \)
• For \( x = 4 \): \( f(4) = -4^3 + 2(4)^2 - 4 = -64 + 32 - 4 = -36 \)

Now, we look for sign changes:

• Between \( x = -2 \) ( \( f(-2) = 12 \) ) and \( x = -1 \) ( \( f(-1) = -1 \) ): Sign changes from positive to negative, so there's a zero between \( -2 \) and \( -1 \).

Problem 3: \( f(x) = x^3 + 4x^2 - 5x + 5 \)

This is a cubic function (degree 3). Let's evaluate \( f(x) \) at integer values of \( x \):

• For \( x = -5 \): \( f(-5) = (-5)^3 + 4(-5)^2 - 5(-5) + 5 = -125 + 100 + 25 + 5 = 5 \)
• For \( x = -4 \): \( f(-4) = (-4)^3 + 4(-4)^2 - 5(-4) + 5 = -64 + 64 + 20 + 5 = 25 \)
• For \( x = -3 \): \( f(-3) = (-3)^3 + 4(-3)^2 - 5(-3) + 5 = -27 + 36 + 15 + 5 = 29 \)
• For \( x = -2 \): \( f(-2) = (-2)^3 + 4(-2)^2 - 5(-2) + 5 = -8 + 16 + 10 + 5 = 23 \)
• For \( x = -1 \): \( f(-1) = (-1)^3 + 4(-1)^2 - 5(-1) + 5 = -1 + 4 + 5 + 5 = 13 \)
• For \( x = 0 \): \( f(0) = 0^3 + 4(0)^2 - 5(0) + 5 = 5 \)
• For \( x = 1 \): \( f(1) = 1^3 + 4(1)^2 - 5(1) + 5 = 1 + 4 - 5 + 5 = 5 \)
• For \( x = 2 \): \( f(2) = 2^3 + 4(2)^2 - 5(2) + 5 = 8 + 16 - 10 + 5 = 19 \)

Wait, all these values are positive? Let's check \( x = -6 \):

• For \( x = -6 \): \( f(-6) = (-6)^3 + 4(-6)^2 - 5(-6) + 5 = -216 + 144 + 30 + 5 = -37 \)

Ah, there we go. So:

• For \( x = -6 \): \( f(-6) = -37 \) (negative)
• For \( x = -5 \): \( f(-5) = 5 \) (positive)

Sign changes from negative to positive between \( x = -6 \) and \( x = -5 \), so there's a zero between \( -6 \) and \( -5 \).

Problem 4: \( f(x) = -x^4 - x^3 + 4 \)

This is a quartic function (degree 4), so it has at most 4 real zeros. Let's evaluate \( f(x) \) at integer values of \( x \):

• For \( x = -3 \): \( f(-3) = -(-3)^4 - (-3)^3 + 4 = -81 + 27 + 4 = -50 \)
• For \( x = -2 \): \( f(-2) = -(-2)^4 - (-2)^3 + 4 = -16 + 8 + 4 = -4 \)
• For \( x = -1 \): \( f(-1) = -(-1)^4 - (-1)^3 + 4 = -1 + 1 + 4 = 4 \)
• For \( x = 0 \): \( f(0) = -0^4 - 0^3 + 4 = 4 \)
• For \( x = 1 \): \( f(1) = -1^4 - 1^3 + 4 = -1 - 1 + 4 = 2 \)
• For \( x = 2 \): \( f(2) = -2^4 - 2^3 + 4 = -16 - 8 + 4 = -20 \)

Now, we look for sign changes:

• Between \( x = -2 \) ( \( f(-2) = -4 \) ) and \( x = -1 \) ( \( f(-1) = 4 \) ): Sign changes from negative to positive, so there's a zero between \( -2 \) and \( -1 \).
• Between \( x = 1 \) ( \( f(1) = 2 \) ) and \( x = 2 \) ( \( f(2) = -20 \) ): Sign changes from positive to negative, so there's a zero between \( 1 \) and \( 2 \).

Final Answers (Zeros Between):

1. \( f(x) = x^2 + 3x - 1 \): Between \( -4 \) and \( -3 \), and between \( 0 \) and \( 1 \).
2. \( f(x) = -x^3 + 2x^2 - 4 \): Between \( -2 \) and \( -1 \).
3. \( f(x) = x^3 + 4x^2 - 5x + 5 \): Between \( -6 \) and \( -5 \).
4. \( f(x) = -x^4 - x^3 + 4 \): Between \( -2 \) and \( -1 \), and between \( 1 \) and \( 2 \).