first derivation = 2 - 3 / x²
second derivation = 6 / x³
Second derivation is undefined when:
6 / x³ →∞
6 / x³ →∞ when x = 0
f''(x) is undefined at x = 0
The graph of a function is concave up on interval where f "(x) > 0
The graph of a function is concave down on interval where y = f "(x) < 0
The graph of a function is concave up when:
f "(x) > 0
6 / x³ > 0
x ∈ ( 0 , ∞ )
The graph of a function is concave down when:
6 / x³ < 0
x ∈ ( - ∞ , 0 )
Consider the function f(x) = 2 x + 3 x ^ { -1 }.
Note that this function has no inflection points, but f''(x) is undefined at x=B where
B=
6/x^3(my answer)
For each of the following intervals, tell whether f(x) is concave up (type in CU) or concave down (type in CD).
(-\infty, B):
CU(my answer)
(B,\infty):
CD(my answer)
hello, my answer in inccorect but i don't know which question is wrong
2 answers
y = 2 x + 3/x Undefined at x = 0
dy/dx = 2 - 3/x^2 also undefined at x = 0
d^2y/dx^2 = +6x/x^4 = 6/x^3
when x is <0 which means <B
d^2/dy^2 <0 so sheds water, concave down
when x>0
d^2y/dx^2 >0 so holds water, concave up
to check definitions use
y = x^2 which we know holds water
dy/dx = 2x
d^2y/dx^2 = +2
dy/dx = 2 - 3/x^2 also undefined at x = 0
d^2y/dx^2 = +6x/x^4 = 6/x^3
when x is <0 which means <B
d^2/dy^2 <0 so sheds water, concave down
when x>0
d^2y/dx^2 >0 so holds water, concave up
to check definitions use
y = x^2 which we know holds water
dy/dx = 2x
d^2y/dx^2 = +2