Solved on Feb 26, 2024

Determine if $f(x) = x^3 + 5x^2 - 8x - 16$ has a real zero between $a = -7$ and $b = -3$ using the intermediate value theorem. A. $f(a) = \square$ and $f(b) = \square$ show the function has at least one real zero. B. $f(a) = \square$ and $f(b) = \square$ show the function does not have a real zero. C. It is impossible to use the intermediate value theorem in this case.

STEP 1

Assumptions
1. The function given is $f(x) = x^{3} + 5x^{2} - 8x - 16$ .
2. We are given two points, $a = -7$ and $b = -3$ .
3. The Intermediate Value Theorem states that if $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there exists at least one number $c$ in the open interval $(a, b)$ such that $f(c) = N$ .

STEP 2

First, we need to verify that the function $f(x)$ is continuous on the interval $[a, b]$ . Since $f(x)$ is a polynomial function, it is continuous everywhere, including on $[a, b]$ .

STEP 3

Next, we will calculate $f(a)$ by substituting $x = a$ into the function $f(x)$ .
$f(a) = f(-7) = (-7)^{3} + 5(-7)^{2} - 8(-7) - 16$

STEP 4

Calculate $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16$

STEP 5

Simplify $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16 = -58$

STEP 6

Now, we will calculate $f(b)$ by substituting $x = b$ into the function $f(x)$ .
$f(b) = f(-3) = (-3)^{3} + 5(-3)^{2} - 8(-3) - 16$

STEP 7

Calculate $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16$

STEP 8

Simplify $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16 = 26$

STEP 9

Now that we have $f(a)$ and $f(b)$ , we can apply the Intermediate Value Theorem. We see that $f(a)$ is negative and $f(b)$ is positive, which means there must be at least one real zero between $a$ and $b$ because the continuous function $f(x)$ must cross the x-axis at least once when it goes from a negative value to a positive value.
Therefore, the correct choice is A, and we fill in the answer boxes with the calculated values of $f(a)$ and $f(b)$ .
A. By the intermediate value theorem, the function has at least one real zero between $a$ and $b$ because $f(a) = -58$ and $f(b) = 26$ .

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STEP 1

STEP 2

First, we need to verify that the function $f(x)$ is continuous on the interval $[a, b]$ . Since $f(x)$ is a polynomial function, it is continuous everywhere, including on $[a, b]$ .

STEP 3

Next, we will calculate $f(a)$ by substituting $x = a$ into the function $f(x)$ .
$f(a) = f(-7) = (-7)^{3} + 5(-7)^{2} - 8(-7) - 16$

STEP 4

Calculate $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16$

STEP 5

Simplify $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16 = -58$

STEP 6

Now, we will calculate $f(b)$ by substituting $x = b$ into the function $f(x)$ .
$f(b) = f(-3) = (-3)^{3} + 5(-3)^{2} - 8(-3) - 16$

STEP 7

Calculate $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16$

STEP 8

Simplify $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16 = 26$

STEP 9

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STEP 1

STEP 2

First, we need to verify that the function f(x) f(x) f(x) is continuous on the interval [a,b][a, b][a,b]. Since f(x) f(x) f(x) is a polynomial function, it is continuous everywhere, including on [a,b][a, b][a,b].

STEP 3

Next, we will calculate f(a) f(a) f(a) by substituting x=a x = a x=a into the function f(x) f(x) f(x).f(a)=f(−7)=(−7)3+5(−7)2−8(−7)−16 f(a) = f(-7) = (-7)^{3} + 5(-7)^{2} - 8(-7) - 16 f(a)=f(−7)=(−7)3+5(−7)2−8(−7)−16

STEP 4

Calculate f(a) f(a) f(a).f(a)=f(−7)=−343+245+56−16 f(a) = f(-7) = -343 + 245 + 56 - 16 f(a)=f(−7)=−343+245+56−16

STEP 5

Simplify f(a) f(a) f(a).f(a)=f(−7)=−343+245+56−16=−58 f(a) = f(-7) = -343 + 245 + 56 - 16 = -58 f(a)=f(−7)=−343+245+56−16=−58

STEP 6

Now, we will calculate f(b) f(b) f(b) by substituting x=b x = b x=b into the function f(x) f(x) f(x).f(b)=f(−3)=(−3)3+5(−3)2−8(−3)−16 f(b) = f(-3) = (-3)^{3} + 5(-3)^{2} - 8(-3) - 16 f(b)=f(−3)=(−3)3+5(−3)2−8(−3)−16

STEP 7

Calculate f(b) f(b) f(b).f(b)=f(−3)=−27+45+24−16 f(b) = f(-3) = -27 + 45 + 24 - 16 f(b)=f(−3)=−27+45+24−16

STEP 8

Simplify f(b) f(b) f(b).f(b)=f(−3)=−27+45+24−16=26 f(b) = f(-3) = -27 + 45 + 24 - 16 = 26 f(b)=f(−3)=−27+45+24−16=26

STEP 9

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First, we need to verify that the function $f(x)$ is continuous on the interval $[a, b]$ . Since $f(x)$ is a polynomial function, it is continuous everywhere, including on $[a, b]$ .

Next, we will calculate $f(a)$ by substituting $x = a$ into the function $f(x)$ .
$f(a) = f(-7) = (-7)^{3} + 5(-7)^{2} - 8(-7) - 16$

Calculate $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16$

Simplify $f(a)$ .
$f(a) = f(-7) = -343 + 245 + 56 - 16 = -58$

Now, we will calculate $f(b)$ by substituting $x = b$ into the function $f(x)$ .
$f(b) = f(-3) = (-3)^{3} + 5(-3)^{2} - 8(-3) - 16$

Calculate $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16$

Simplify $f(b)$ .
$f(b) = f(-3) = -27 + 45 + 24 - 16 = 26$