𝑓′(𝑥) < 0 for xe (−∞, 1) ∪ (5, ∞)
𝑓′′(𝑥) > 0 for xe (−1,3) ∪ (6, ∞)
lim x→3^+ 𝑓(𝑥) = −∞ , (Note: as x approaches 3 from above, f(x) = −∞)
lim 𝑥→±∞ 𝑓(𝑥) = 0, (Note: as x approaches positive and negative infinity, 𝑓(𝑥) = 0)
𝑓′′(−1) = 𝑓′(5) = 0
how would function 𝑓(𝑥) be drawn to satisfy the properties above.
To sketch a function that satisfies the given properties, we can proceed as follows:
1. First, let's analyze the information given about the derivative, 𝑓′(𝑥):
- 𝑓′(𝑥) < 0 for 𝑥 ∈ (−∞, 1) ∪ (5, ∞). This means that the function is decreasing in the intervals 𝑥 < 1 and 𝑥 > 5.
- 𝑓′′(𝑥) > 0 for 𝑥 ∈ (−1,3) ∪ (6, ∞). This indicates that the function is concave up in the intervals 𝑥 > −1 and 𝑥 > 6.
- 𝑓′′(−1) = 𝑓′(5) = 0. This tells us that the function has an inflection point at 𝑥 = −1 and 𝑥 = 5.
2. To satisfy the given limits:
- lim 𝑥→3^+ 𝑓(𝑥) = −∞. As 𝑥 approaches 3 from above, 𝑓(𝑥) should decrease without bound, indicating a vertical asymptote at 𝑥 = 3.
- lim 𝑥→±∞ 𝑓(𝑥) = 0. As 𝑥 approaches positive and negative infinity, 𝑓(𝑥) should tend towards zero, suggesting horizontal asymptotes at 𝑦 = 0.
Based on this information, we can sketch the function 𝑓(𝑥) as follows:
- Start by drawing a decreasing curve in the intervals 𝑥 < 1 and 𝑥 > 5. The function should have a local maximum at 𝑥 = 1 and a local minimum at 𝑥 = 5.
- Draw an inflection point at 𝑥 = −1 and 𝑥 = 5, ensuring that the concavity changes from concave down to concave up at 𝑥 = −1 and remains concave up after 𝑥 = 5.
- As 𝑥 approaches 3 from above, draw a vertical asymptote at 𝑥 = 3, indicating that 𝑓(𝑥) decreases without bound.
- Finally, ensure that as 𝑥 approaches positive and negative infinity, 𝑓(𝑥) tends towards the horizontal asymptote 𝑦 = 0.
Remember that this is just one potential sketch that satisfies the given properties. There may be multiple ways to draw a function that satisfies these conditions.