Asked by guy022
f(x) = (x^2 - 1) ^3
a) domain
b) intercepts
c) asymptotes
d) symmetry
e) critical values of the 1st derivative
f) intervals of increase/decrease
g) local and extreme maximum/minimum points
h) critical values of the 2nd derivative
i) concavity
j) points of inflection
a) domain
b) intercepts
c) asymptotes
d) symmetry
e) critical values of the 1st derivative
f) intervals of increase/decrease
g) local and extreme maximum/minimum points
h) critical values of the 2nd derivative
i) concavity
j) points of inflection
Answers
Answered by
Help
I got the derivative for e but not sure how to pull out the critical values from 6x^5 -12x^3 + 6x and then the intervals, locals and extremes always confuses me
Answered by
oobleck
(a) domain: all reals (as with any polynomial)
(b) x = ±1
(c) no asymptotes (as with any polynomial)
(d) y-axis
(e) f' = 6x(x^2-1)^2, so f'=0 at x = -1, 0, 1
(f) f' > 0 for x > 0 (except at x=1, where f' = 0)
f' < 0 for x < 0 (except at x=-1, where f' = 0)
(g) f(0) = -1
(h) f" = 6(x^2-1)(5x^2-1) so f"=0 at x = ±1, ±1/√5
(i) f" > 0 (concave up) for x < -1, -1/√5 < x < 1/√5, x > 1
down elsewhere
(j) where f"=0
(b) x = ±1
(c) no asymptotes (as with any polynomial)
(d) y-axis
(e) f' = 6x(x^2-1)^2, so f'=0 at x = -1, 0, 1
(f) f' > 0 for x > 0 (except at x=1, where f' = 0)
f' < 0 for x < 0 (except at x=-1, where f' = 0)
(g) f(0) = -1
(h) f" = 6(x^2-1)(5x^2-1) so f"=0 at x = ±1, ±1/√5
(i) f" > 0 (concave up) for x < -1, -1/√5 < x < 1/√5, x > 1
down elsewhere
(j) where f"=0
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