91Ó°ÊÓ

To find the critical points, we will first find the derivative of the function. The function is given by \(f(x) = \frac{2x^2 - 24}{x + 4}\). To find the derivative, use the quotient rule, given by \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\), where \(u\) and \(v\) are differentiable functions. In this case, \(u(x) = 2x^2 - 24\) and \(v(x) = x + 4\). We begin by finding the derivatives of \(u\) and \(v\): \(u'(x) = 4x\) \(v'(x) = 1\) Use the quotient rule to find the derivative of \(f(x)\): \(f'(x) = \frac{(x + 4)(4x) - (2x^2 - 24)(1)}{(x + 4)^2}\) Now simplify the derivative: \(f'(x) = \frac{4x^2 + 16x - 2x^2 + 24}{(x + 4)^2}\) \(f'(x) = \frac{2x^2 + 16x + 24}{(x + 4)^2}\)

To find the critical points, set the derivative equal to zero, and find where the derivative is undefined: \(0 = \frac{2x^2 + 16x + 24}{(x + 4)^2}\) The derivative is undefined when the denominator is equal to zero, which occurs at \(x = -4\). Now, let's solve for when the numerator is equal to zero: \(0 = 2x^2 + 16x + 24\) Divide both sides by 2: \(0 = x^2 + 8x + 12\) Factor the quadratic: \(0 = (x + 2)(x + 6)\) Solve for x: \(x = -2, -6\) Thus, there are three critical points: \(x = -4\), \(x = -2\), and \(x = -6\).

To determine whether the critical points correspond to extrema, we will use the second derivative test. First, let's find the second derivative of the function: \(f'(x) = \frac{2x^2 + 16x + 24}{(x + 4)^2}\) Applying quotient rule for second derivative, \(f''(x) = \frac{((x + 4)^2)(4x + 16) - (2x^2 + 16x + 24)(2(x + 4))}{(x + 4)^4}\) Now evaluate the second derivative at the critical points: \(f''(-4) = \frac{(-4 + 4)^2(4(-4) + 16) - (2(-4)^2 + 16(-4) + 24)(2(-4 + 4))}{(-4 + 4)^4} = \) Undefined \(f''(-2) = \frac{(-2 + 4)^2(4(-2) + 16) - (2(-2)^2 + 16(-2) + 24)(2(-2 + 4))}{(-2 + 4)^4} = \frac{8}{16} > 0\) \(f''(-6) = \frac{(-6 + 4)^2(4(-6) + 16) - (2(-6)^2 + 16(-6) + 24)(2(-6 + 4))}{(-6 + 4)^4} = -\frac{32}{16} < 0\) The second derivative test states that if the second derivative is positive at a critical point, the point is a relative minimum. If it's negative, the point is a relative maximum. In this case: - At point x = -2, the second derivative is positive, indicating a relative minimum. - At point x = -6, the second derivative is negative, indicating a relative maximum. So we have a relative minimum at x = -2 and a relative maximum at x = -6. To find their locations, we need to evaluate the original function at these points: \(f(-2) = \frac{2(-2)^2 - 24}{-2 + 4} = 4\) \(f(-6) = \frac{2(-6)^2 - 24}{-6 + 4} = -12\) Thus, the relative minimum is at the point \((-2, 4)\), and the relative maximum is at the point \((-6, -12)\).

Since there are no endpoints of the domain, and we have already found the relative extrema, there are no additional points to check for absolute extrema. Consequently, the relative extrema found in the previous step are also the absolute extrema. The absolute minimum is \((-2, 4)\), and the absolute maximum is \((-6, -12)\).