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Find the two [tex]\( x \)[/tex]-intercepts of the function [tex]\( f \)[/tex] and show that [tex]\( f^{\prime}(x) = 0 \)[/tex] at some point between the two [tex]\( x \)[/tex]-intercepts.

[tex]\[ f(x) = x^2 - 2x - 15 \][/tex]

[tex]\[
\begin{array}{c}
(x, y) = (\square) \text{ (smaller } x \text{-value)} \\
(x, y) = (\square) \text{ (larger } x \text{-value)}
\end{array}
\][/tex]

Find a value of [tex]\( x \)[/tex] such that [tex]\( f^{\prime}(x) = 0 \)[/tex].

[tex]\[ x = \square \][/tex]

Sagot :

GinnyAnswer
Let's solve this problem step-by-step.

We are given the function:
[tex]\[ f(x) = x^2 - 2x - 15 \][/tex]

### 1. Finding the x-intercepts:
To find the x-intercepts, we need to solve the equation [tex]\( f(x) = 0 \)[/tex]. This means we solve:
[tex]\[ x^2 - 2x - 15 = 0 \][/tex]

This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. We can solve it by factoring or using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].

Let's factor this quadratic equation. We look for two numbers that multiply to [tex]\(-15\)[/tex] (the constant term) and add to [tex]\(-2\)[/tex] (the coefficient of the linear term). The numbers [tex]\(-5\)[/tex] and [tex]\(3\)[/tex] satisfy these conditions.

Thus, we can write:
[tex]\[ x^2 - 2x - 15 = (x - 5)(x + 3) = 0 \][/tex]

Setting each factor equal to zero gives us the x-intercepts:
[tex]\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \][/tex]
[tex]\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \][/tex]

So, the two x-intercepts are [tex]\( (-3, 0) \)[/tex] (smaller [tex]\( x \)[/tex]-value) and [tex]\( (5, 0) \)[/tex] (larger [tex]\( x \)[/tex]-value).

### 2. Showing that [tex]\( f'(x) = 0 \)[/tex] at some point between the two x-intercepts:
Next, we need to find the derivative of the function [tex]\( f(x) \)[/tex]. The derivative [tex]\( f'(x) \)[/tex] gives us the slope of the function at any point [tex]\( x \)[/tex].

[tex]\[ f(x) = x^2 - 2x - 15 \][/tex]
[tex]\[ f'(x) = \frac{d}{dx} (x^2 - 2x - 15) = 2x - 2 \][/tex]

We need to find the value of [tex]\( x \)[/tex] where the derivative [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 2x - 2 = 0 \][/tex]
[tex]\[ 2x = 2 \][/tex]
[tex]\[ x = 1 \][/tex]

So, [tex]\( f'(x) = 0 \)[/tex] at [tex]\( x = 1 \)[/tex]. This point [tex]\( x = 1 \)[/tex] is indeed between the two x-intercepts [tex]\( -3 \)[/tex] and [tex]\( 5 \)[/tex].

### Summary:
[tex]\[ \begin{array}{c} f(x) = x^2 - 2x - 15 \\ (x, y) = (-3, 0) \quad \text{(smaller } x \text{-value)} \\ (x, y) = (5, 0) \quad \text{(larger } x \text{-value)} \end{array} \][/tex]

The value of [tex]\( x \)[/tex] such that [tex]\( f'(x) = 0 \)[/tex] is:
[tex]\[ x = 1 \][/tex]

Thus, both the x-intercepts and the critical point have been identified.
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