Answer:
To find the relative maxima, relative minima, and horizontal points of inflection for the given function \(y = \frac{1}{3}x^3 + x^2 - 8x + 10\), we'll need to calculate its first and second derivatives and then analyze their critical points. Let's start with finding the first and second derivatives:
1. Find the first derivative \(y'\):
\(y = \frac{1}{3}x^3 + x^2 - 8x + 10\)
\(y' = \frac{d}{dx}\left(\frac{1}{3}x^3 + x^2 - 8x + 10\right)\)
\(y' = x^2 + 2x - 8\)
2. Find the second derivative \(y''\):
\(y'' = \frac{d}{dx}(x^2 + 2x - 8)\)
\(y'' = 2x + 2\)
Now, we'll find the critical points by setting \(y' = 0\) and solving for \(x\):
\(x^2 + 2x - 8 = 0\)
This is a quadratic equation, and we can solve it by factoring:
\((x + 4)(x - 2) = 0\)
So, \(x = -4\) and \(x = 2\) are the critical points.
Now, let's analyze these critical points to find the relative maxima, relative minima, and horizontal points of inflection:
1. Relative Maxima/Minima:
To determine whether these critical points are relative maxima or minima, we can use the second derivative test. Plug these critical points into the second derivative \(y''\):
For \(x = -4\), \(y''(-4) = 2(-4) + 2 = -8 + 2 = -6\)
Since \(y''(-4)\) is negative, this implies a relative maximum at \((x, y) = (-4, y(-4))\).
For \(x = 2\), \(y''(2) = 2(2) + 2 = 4 + 2 = 6\)
Since \(y''(2)\) is positive, this implies a relative minimum at \((x, y) = (2, y(2))\).
2. Horizontal Point of Inflection:
To find the horizontal point of inflection, we need to find where the second derivative \(y''\) equals zero. However, since \(y''\) is a linear function (\(y'' = 2x + 2\)), it never equals zero. Therefore, there are no horizontal points of inflection for this function.
So, in summary:
- Relative Maxima: \((x, y) = (-4, y(-4)) = (-4, 36)\)
- Relative Minima: \((x, y) = (2, y(2)) = (2, 0.67)\)
- Horizontal Points of Inflection: DNE (None exist)
Explanation: