Certainly! Let's address each part of the question in a detailed step-by-step manner.
### Part (a)
We are asked to show that the equation [tex]\( x^3 + 6x - 5 = 0 \)[/tex] has a solution between [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex].
1. Evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 0^3 + 6 \cdot 0 - 5 = -5
\][/tex]
2. Evaluate the function at [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 1^3 + 6 \cdot 1 - 5 = 1 + 6 - 5 = 2
\][/tex]
3. Analyze the signs of [tex]\( f(0) \)[/tex] and [tex]\( f(1) \)[/tex]:
[tex]\[
f(0) = -5 \quad \text{(negative)}, \quad f(1) = 2 \quad \text{(positive)}
\][/tex]
Given that [tex]\( f(0) \)[/tex] is negative and [tex]\( f(1) \)[/tex] is positive, and since [tex]\( f(x) \)[/tex] is a continuous function (as it is a polynomial), by the Intermediate Value Theorem, there must be at least one root of the equation [tex]\( x^3 + 6x - 5 = 0 \)[/tex] in the interval [tex]\( (0, 1) \)[/tex].
Thus, the equation [tex]\( x^3 + 6x - 5 = 0 \)[/tex] has a solution between [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex].
### Part (b)
We need to show how the equation [tex]\( x^3 + 6x - 5 = 0 \)[/tex] can be rearranged to give [tex]\( x = \frac{5}{x^2 + 6} \)[/tex] by filling in the boxes in the provided template.
1. Starting with the given equation:
[tex]\[
x^3 + 6x - 5 = 0
\][/tex]
2. Move the constant term to the right-hand side:
[tex]\[
x^3 + 6x = 5
\][/tex]
3. Express [tex]\( x \)[/tex] in terms of the remaining expression:
[tex]\[
x = \frac{5}{x^2 + 6}
\][/tex]
We can now fill in the boxes:
- For the template:
[tex]\[
x^3 + 6x = \square
\][/tex]
The filled answer is:
[tex]\[
x^3 + 6x = 5
\][/tex]
- For the template:
[tex]\[
\square \left(x^2 + 6\right) = x
\][/tex]
Rearranging [tex]\( x = \frac{5}{x^2 + 6} \)[/tex] to [tex]\(5 = x \cdot \left( x^2 + 6 \right)\)[/tex], the filled answer is:
[tex]\[
5 = x \left(x^2 + 6\right)
\][/tex]
Thus, we have shown that the equation [tex]\( x^3 + 6x - 5 = 0 \)[/tex] can indeed be rearranged into [tex]\( x = \frac{5}{x^2 + 6} \)[/tex].
