this question has two parts. first, answer part a. then, answer part b.
part a
determine the consecutive integer values of x between which each
real zero of $f(x) = 2x^4 + x^3 - 3x^2 - 2$ is located.
part a
select all intervals in which a real zero is located.
a) $x = -4$ and $x = -3$
b) $x = -3$ and $x = -2$
c) $x = -2$ and $x = -1$
d) $x = -1$ and $x = 0$
e) $x = 0$ and $x = 1$
f) $x = 1$ and $x = 2$

Question

Accepted Answer

# Explanation:
To determine the intervals where the real zeros of $ f(x) = 2x^4 + x^3 - 3x^2 - 2 $ are located, we can use the Intermediate Value Theorem. The Intermediate Value Theorem 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 exists at least one real zero of $ f(x) $ in the interval $(a, b)$.

First, we calculate $ f(x) $ at the integer values of $ x $ from $-4$ to $ 2 $:

## Step 1: Calculate $ f(-4) $
$$
\begin{align*}
f(-4) &= 2(-4)^4 + (-4)^3 - 3(-4)^2 - 2 \
&= 2(256) + (-64) - 3(16) - 2 \
&= 512 - 64 - 48 - 2 \
&= 398
\end{align*}
$$

## Step 2: Calculate $ f(-3) $
$$
\begin{align*}
f(-3) &= 2(-3)^4 + (-3)^3 - 3(-3)^2 - 2 \
&= 2(81) + (-27) - 3(9) - 2 \
&= 162 - 27 - 27 - 2 \
&= 106
\end{align*}
$$

## Step 3: Calculate $ f(-2) $
$$
\begin{align*}
f(-2) &= 2(-2)^4 + (-2)^3 - 3(-2)^2 - 2 \
&= 2(16) + (-8) - 3(4) - 2 \
&= 32 - 8 - 12 - 2 \
&= 10
\end{align*}
$$

## Step 4: Calculate $ f(-1) $
$$
\begin{align*}
f(-1) &= 2(-1)^4 + (-1)^3 - 3(-1)^2 - 2 \
&= 2(1) + (-1) - 3(1) - 2 \
&= 2 - 1 - 3 - 2 \
&= -4
\end{align*}
$$

## Step 5: Calculate $ f(0) $
$$
\begin{align*}
f(0) &= 2(0)^4 + (0)^3 - 3(0)^2 - 2 \
&= 0 + 0 - 0 - 2 \
&= -2
\end{align*}
$$

## Step 6: Calculate $ f(1) $
$$
\begin{align*}
f(1) &= 2(1)^4 + (1)^3 - 3(1)^2 - 2 \
&= 2(1) + 1 - 3(1) - 2 \
&= 2 + 1 - 3 - 2 \
&= -2
\end{align*}
$$

## Step 7: Calculate $ f(2) $
$$
\begin{align*}
f(2) &= 2(2)^4 + (2)^3 - 3(2)^2 - 2 \
&= 2(16) + 8 - 3(4) - 2 \
&= 32 + 8 - 12 - 2 \
&= 26
\end{align*}
$$

Now, we check the sign changes between consecutive integers:

- Between $ x = -2 $ and $ x = -1 $: $ f(-2) = 10 $ (positive) and $ f(-1) = -4 $ (negative). There is a sign change, so a zero is in $(-2, -1)$.
- Between $ x = 1 $ and $ x = 2 $: $ f(1) = -2 $ (negative) and $ f(2) = 26 $ (positive). There is a sign change, so a zero is in $(1, 2)$.

For the other intervals:

- $ x = -4 $ to $ x = -3 $: Both $ f(-4) $ and $ f(-3) $ are positive. No sign change.
- $ x = -3 $ to $ x = -2 $: Both $ f(-3) $ and $ f(-2) $ are positive. No sign change.
- $ x = -1 $ to $ x = 0 $: Both $ f(-1) $ and $ f(0) $ are negative. No sign change.
- $ x = 0 $ to $ x = 1 $: Both $ f(0) $ and $ f(1) $ are negative. No sign change.

# Answer:
C) $ x = -2 $ and $ x = -1 $
F) $ x = 1 $ and $ x = 2 $

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QUESTION IMAGE

Question

Explanation:

Step 1: Calculate \( f(-4) \)

Step 2: Calculate \( f(-3) \)

Step 3: Calculate \( f(-2) \)

Step 4: Calculate \( f(-1) \)

Step 5: Calculate \( f(0) \)

Step 6: Calculate \( f(1) \)

Step 7: Calculate \( f(2) \)

Answer: