Certainly! Let's solve this problem step-by-step:
1. Given Function:
[tex]\[
y = \frac{1}{3} x^3 + 2.5 x^2 + 6 x - 5
\][/tex]
2. Find the First Derivative [tex]\(\frac{dy}{dx}\)[/tex]:
We need to calculate the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex].
[tex]\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{3} x^3 + 2.5 x^2 + 6 x - 5 \right)
\][/tex]
By applying basic differentiation rules to each term separately, we get:
[tex]\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{3} x^3 \right) + \frac{d}{dx} \left( 2.5 x^2 \right) + \frac{d}{dx} \left( 6 x \right) + \frac{d}{dx} \left( -5 \right)
\][/tex]
Compute the derivatives:
[tex]\[
\frac{d}{dx} \left( \frac{1}{3} x^3 \right) = x^2
\][/tex]
[tex]\[
\frac{d}{dx} \left(2.5 x^2 \right) = 5x
\][/tex]
[tex]\[
\frac{d}{dx} \left(6 x \right) = 6
\][/tex]
[tex]\[
\frac{d}{dx} \left(-5\right) = 0
\][/tex]
So, combining these results:
[tex]\[
\frac{dy}{dx} = x^2 + 5x + 6
\][/tex]
3. Set the Derivative Equal to Zero:
To find the critical points, where the slope of the function is zero, we set the derivative equal to zero:
[tex]\[
x^2 + 5x + 6 = 0
\][/tex]
4. Solve the Quadratic Equation:
We solve the quadratic equation [tex]\(x^2 + 5x + 6 = 0\)[/tex] using the factorization method or the quadratic formula [tex]\(\left( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \right)\)[/tex].
First, let's factorize it:
[tex]\[
x^2 + 5x + 6 = (x + 3)(x + 2) = 0
\][/tex]
Setting each bracket to zero gives us the solutions:
[tex]\[
x + 3 = 0 \quad \text{or} \quad x + 2 = 0
\][/tex]
Solving these:
[tex]\[
x = -3 \quad \text{or} \quad x = -2
\][/tex]
5. Final Result:
The values of [tex]\(x\)[/tex] when [tex]\(\frac{dy}{dx} = 0\)[/tex] are:
[tex]\[
x = -3 \quad \text{and} \quad x = -2
\][/tex]
These are the points where the slope of the function [tex]\(y\)[/tex] is zero.
