- Calculate the first and second derivatives of \(f(x) = e^{5 - 50x^2}\) to find \(f''(x) = 100 e^{5 - 50x^2}(100x^2 - 1)\).
- Determine the intervals of concavity by testing the sign of \(f''(x)\) around its roots at \(x = \pm 0.1\).
- Identify that the function is concave up on \((-\infty, -0.1) \cup (0.1, \infty)\) and concave down on \((-0.1, 0.1)\).
- State the x-values of the inflection points as \(x = -0.1, 0.1\).
- Final Answer: (a) \(\boxed{(-\infty, -0.1) \cup (0.1, \infty)}\); (b) \(\boxed{(-0.1, 0.1)}\); (c) \(\boxed{x = -0.1, 0.1}\)
Explanation
- Calculating the Derivatives
To determine the concavity and inflection points of the function \(f(x) = e^{5 - 50x^2}\), we analyze its derivatives. The first derivative \(f'(x)\) is found using the chain rule:
\[f'(x) = e^{5 - 50x^2} \cdot \frac{d}{dx}(5 - 50x^2) = -100x e^{5 - 50x^2}\]
Next, we calculate the second derivative \(f''(x)\) using the product rule on \(f'(x)\):
\[f''(x) = \frac{d}{dx}(-100x) \cdot e^{5 - 50x^2} + (-100x) \cdot \frac{d}{dx}(e^{5 - 50x^2})\]
\[f''(x) = -100 e^{5 - 50x^2} + (-100x) \cdot (-100x e^{5 - 50x^2})\]
\[f''(x) = -100 e^{5 - 50x^2} + 10000x^2 e^{5 - 50x^2} = 100 e^{5 - 50x^2}(100x^2 - 1)\]
- Analyzing Concavity
The function is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\). Since the term \(100 e^{5 - 50x^2}\) is always positive, the sign of \(f''(x)\) depends solely on the quadratic expression \(100x^2 - 1\). The roots of \(100x^2 - 1 = 0\) are \(x = \pm 0.1\).
- For \(x \in (-\infty, -0.1)\), choosing \(x = -1\): \(100(-1)^2 - 1 = 99 > 0\).
- For \(x \in (-0.1, 0.1)\), choosing \(x = 0\): \(100(0)^2 - 1 = -1 < 0\).
- For \(x \in (0.1, \infty)\), choosing \(x = 1\): \(100(1)^2 - 1 = 99 > 0\).
Thus, the function is concave up on \((-\infty, -0.1) \cup (0.1, \infty)\) and concave down on \((-0.1, 0.1)\).
- Identifying Inflection Points
Inflection points occur where the second derivative changes sign. Based on our analysis in the previous step, the sign of \(f''(x)\) changes at \(x = -0.1\) and \(x = 0.1\). Therefore, these points are the x-values of the inflection points.
- Final Answer
AConcave up: \((-\infty, -0.1) \cup (0.1, \infty)\)
BConcave down: \((-0.1, 0.1)\)
CInflection points: \(x = -0.1, 0.1\)
Examples
Understanding concavity and inflection points is essential in fields like physics and engineering, particularly when analyzing the acceleration of a moving object. For instance, if you are tracking the position of a vehicle over time, the function's concavity tells you whether the velocity is increasing or decreasing. Inflection points represent the exact moments when the vehicle transitions from speeding up to slowing down (or vice versa), allowing engineers to optimize safety features and performance metrics in automotive design.