f(x) = \frac{ x^3 }{ x^2 - 25 }
defined on the interval [ -18, 18 ].
Enter points, such as inflection points in ascending order, i.e. smallest x values first. Enter intervals in ascending order also.
The function f(x) has vertical asympototes at (? )and (?) .
f(x) is concave up on the region (?)to (?)and (?) to(?) .
The inflection point for this function is (?)
2 answers
actually i solved my vertical asymptote it's -5 and 5. but how will i find the concave up and inflection point
recall your second derivative
concave up if f" > 0
inflection when f" = 0
concave down if f" < 0
So, you have
f"(x) = 50x(x^2+75)/(x^2-25)^3
f" < 0 on (-∞,-5)
f" > 0 on (-5,0)
f"(0) = 0
f" < 0 on (0,5)
f" > 0 on (5,∞)
So, match that up with the graph of f(x):
http://www.wolframalpha.com/input/?i=x^3%2F%28x^2-25%29
concave up if f" > 0
inflection when f" = 0
concave down if f" < 0
So, you have
f"(x) = 50x(x^2+75)/(x^2-25)^3
f" < 0 on (-∞,-5)
f" > 0 on (-5,0)
f"(0) = 0
f" < 0 on (0,5)
f" > 0 on (5,∞)
So, match that up with the graph of f(x):
http://www.wolframalpha.com/input/?i=x^3%2F%28x^2-25%29
1 answer