Question

f (x) = x^6 - 3x^4 + 3x^2 - 1

Find the:
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


* if you can please show the steps, so that I can follow everything

Answer 1

(a) the domain of ALL polynomials is (-∞,∞)
(b) recall the 3rd row of Pascal's Triangle. f(x) = (x^2-1)^3
so the x-intercepts (roots) are x = ±1
(c) no polynomial has any asymptotes
(d) since f(-x) = f(x), f is an even function, with x=0 as the axis of symmetry
(e) f'(x) = 6x(x^2-1)^2
critical values are thus x = 0, ±1
(f) increasing where f' > 0 ... x > 0
decreasing where f' < 0 ... x < 0
(g) f'=0 at x=0, x = ±1
f" = 6(x^2-1)(5x^2-1)
f"(0) > 0 so f has a minimum there
(h) f" is also zero at x = ±1, so those are not extrema
f" > 0 on (-∞,-1)U(-1/√5,1/√5)U(1,∞) so concave up
f" < 0 on (-1,-1/√5)U(1/√5,1) so concave down
(i) f"=0 at x = ±1, ±1/√5

Answer 2

f (x) = x^6 - 3x^4 + 3x^2 - 1
a) since this is a standard polynomial expression the domain is
the set of real numbers.
b) for y-intercept, let x = 0, and you get (0,-1)
for the x-intercept, we need the x's at which
x^6 - 3x^4 + 3x^2 - 1 = 0
notice that the coefficients add up to zero, so x = 1 is a solution
I divided to get
(x-1)(x^5 + x^4 - 2x^3 - 2x^2 + x + 1) = 0
again we see that the coefficients add up to zero, so x = 1 is a solution
(x-1)(x-1)(x^4 + 2x^3 -2x - 1) = 0
well, how about that ...
(x-1)(x-1)(x-1)(x^3 + 3x^2 + 3x + 1) = 0
and I recognize the last part as (x+1)^3 = x^3+3x^2+3x+1 from
Pascal's triangle

so finally we get
(x-1)^3 (x+1)^3) = 0
so the x intercepts are (1,0) and (-1,0)

c) obviously no asymptotes

d) all the exponents on our variables are even , so there is symmetry
about the y-axis

e) f ' (x) = 6x^5 - 12x^3 + 6x
= 0 for critical values
6x(x^4 - 2x^2 + 1) = 0
6x(x^2 - 1 )^2 = 0
x = 0, x = ± 1

your turn to continue
you should know that f ''(x) = 0 yields the points of inflection
if f '(x) is positive, the function is increasing,
if f '(x) is negative, .......... decreasing

if f ''(x) is positive, function is concave up
if f "(x) is negative, concave down

carry on