91Ó°ÊÓ

Finding the first derivative of a function is the first step in identifying critical points. For the function \( f(x) = \frac{x^{2}}{x+1} \), we apply the quotient rule to get the first derivative. The quotient rule states that for a function \( g(x) = \frac{u(x)}{v(x)} \), the derivative \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). Here, \(u(x) = x^2\) and \(v(x) = x + 1\). So, we compute:
\[ f'(x) = \frac{(x+1) \frac{d}{dx} [x^2] - x^2 \frac{d}{dx} [x+1]}{(x+1)^2} = \frac{(x+1) (2x) - x^2 (1)}{(x+1)^2}. \] Simplifying, we get:\[ f'(x) = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}. \] Critical points occur where \( f'(x) = 0 \). Setting the numerator equal to zero gives \(x^2 + 2x = 0\), leading to critical points at \( x = 0 \) and \( x = -2 \).

The second derivative test helps determine if a critical point is a relative maximum, minimum, or neither. For a function \(f(x)\), the second derivative \( f''(x) \) is found by differentiating \( f'(x) \). In our example, where \( f'(x) = \frac{x^2 + 2x}{(x+1)^2} \), we find:
\[ f''(x) = \frac{(x+1)^2(2x+2) - (x^2 + 2x)2(x+1)}{(x+1)^4} \] Simplifying, we get:
\[ f''(x) = \frac{2x^3 + 6x^2 - 2x}{(x+1)^4}. \] To apply the test, evaluate \( f''(x) \) at the critical points. If \( f''(c) > 0 \), \( c \) is a relative minimum; if \( f''(c) < 0 \), \( c \) is a relative maximum. If \( f''(c) = 0 \), the test is inconclusive.

The quotient rule is a fundamental tool in calculus for finding the derivative of a quotient of two functions. If you have a function \( g(x) = \frac{u(x)}{v(x)} \), its derivative is:\[ g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}. \] In our example, with \( f(x) = \frac{x^2}{x+1} \), apply the rule as follows:\( u(x) = x^2 \) and \( v(x) = x+1 \). Their derivatives are \( u'(x) = 2x \) and \( v'(x) = 1 \). Thus, the first derivative is:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} = \frac{(x+1)(2x) - x^2(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}. \]

Relative maxima occur at points where the function reaches a local highest value. If \( f''(c) < 0 \) for a critical point \( c \), \( c \) is a relative maximum. It means the function is concave down at that point. Using the second derivative test for the function \( f(x) \), we found the critical points at \( x = 0 \) and \( x = -2 \), but the second derivative test was inconclusive as both returned zero:

\( f''(0) = 0 \)
\( f''(-2) = 0 \).

So, we cannot confirm if these points are relative maxima using this test. An alternative method, like the first derivative test, should be utilized to investigate further.

Relative minima are points where the function achieves a local lowest value. When \( f''(c) > 0 \) at a critical point \( c \), it indicates a relative minimum. This means the function is concave up at that point. For our function \( f(x) \), the points \( x = 0 \) and \( x = -2 \) were found to be critical points, but evaluating the second derivative at these points gave:
\( f''(0) = 0 \)
and \( f''(-2) = 0 \), making the second derivative test inconclusive in determining relative minima. We would need to apply other methods, such as the first derivative test, to analyze if these critical points are relative minima.