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.