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The function f(x) = (x^2 + x - 6) e^x has many interesting properties that can be found by analyzing its domain, intercepts, asymptotes, symmetry, critical values, intervals of increase/decrease, local/extreme maximum/minimum points, critical values of the second derivative, concavity, and points of inflection.
a) To find the domain of the function, we need to consider any restrictions imposed by the presence of the exponential function e^x. Since e^x is defined for all real numbers, there are no restrictions on the domain caused by this term. Therefore, the domain of f(x) is all real numbers.
b) Intercepts of the function can be found by setting f(x) = 0 and solving the equation. In this case, we have (x^2 + x - 6) e^x = 0. By factoring the quadratic term, we get (x - 2)(x + 3) e^x = 0. We then solve each factor separately: x - 2 = 0 gives x = 2, and x + 3 = 0 gives x = -3. So the x-intercepts of the function are x = 2 and x = -3.
c) Asymptotes can be found by analyzing the behavior of the function as x approaches positive or negative infinity. In this case, as x approaches positive or negative infinity, the exponential term e^x will dominate the quadratic term. Therefore, there are no horizontal asymptotes. However, there may be vertical asymptotes if the function approaches infinity or negative infinity as x approaches certain values. To find vertical asymptotes, we need to equate the exponential term e^x to infinity or negative infinity. In this case, since e^x is always positive, it can never equal negative infinity. Therefore, there are no vertical asymptotes.
d) To determine symmetry, we need to consider whether the function is even or odd. In this case, we can observe that the function f(x) is not symmetric with respect to the y-axis, which implies it is not an even function. Similarly, if we substitute -x into the function, we find that f(-x) ≠ -f(x). Therefore, the function is not odd either. Hence, the function does not possess any symmetry.
e) Critical values of the first derivative can be found by taking the derivative of the function f(x), setting it equal to zero, and solving the resulting equation. The first derivative of f(x) is f'(x) = (x^2 + 3x - 5) e^x. Setting f'(x) = 0 gives (x^2 + 3x - 5) e^x = 0. Since e^x is always positive, we need to solve the quadratic term instead: x^2 + 3x - 5 = 0. By factoring or using the quadratic formula, we find that x = (-3 ± √29) / 2. Hence, the critical values of the first derivative are x = (-3 + √29) / 2 and x = (-3 - √29) / 2.
f) To determine the intervals of increase and decrease, we need to analyze the behavior of the first derivative. Since the first derivative f'(x) is a product of (x^2 + 3x - 5) and e^x, we can use the sign chart to determine the intervals. The quadratic term x^2 + 3x - 5 has roots at (-3 + √29) / 2 and (-3 - √29) / 2, which divide the real number line into three intervals: (-∞, (-3 - √29) / 2), ((-3 - √29) / 2, (-3 + √29) / 2), and ((-3 + √29) / 2, ∞). We can then test a point within each interval to determine the sign of the first derivative. For example, if we consider x = -4, we find that f'(-4) = (-4^2 + 3(-4) - 5) e^-4 ≈ 0.08. Since f'(-4) is positive, we know that the function is increasing in the interval (-∞, (-3 - √29) / 2). Similarly, we can test the intervals ((-3 - √29) / 2, (-3 + √29) / 2) and ((-3 + √29) / 2, ∞) to determine the intervals of increase and decrease.
g) Local and extreme maximum/minimum points can be found by analyzing the critical points of the function. The critical values of the first derivative are (-3 + √29) / 2 and (-3 - √29) / 2. By plugging these values into the original function f(x), we can find the corresponding y-values. These y-values represent the function's local maximum or minimum points.
h) Critical values of the second derivative can be found by taking the derivative of the first derivative f'(x). The second derivative f''(x) can be calculated as f''(x) = ((x - 1)(x + 4) + 8) e^x. To find the critical values, we set f''(x) = 0 and solve the equation. However, in this case, since the equation is a quadratic multiplied by e^x, there are no critical values of the second derivative.
i) To determine the concavity, we need to analyze the sign of the second derivative. Since the second derivative f''(x) = ((x - 1)(x + 4) + 8) e^x does not have any critical values, we can use the sign chart to determine the intervals of concavity. By testing a point within each interval, we can determine whether the function is concave up or concave down.
j) Points of inflection occur where the concavity changes. By analyzing the sign changes in the second derivative, we can identify the points of inflection. In this case, since the second derivative does not change sign, there are no points of inflection for the function f(x) = (x^2 + x - 6) e^x.
By understanding these properties of the function f(x) = (x^2 + x - 6) e^x, we gain insights into its behavior, allowing us to analyze its graph and make predictions about its features.
