Suppose that f(x)=3/(x^2-16). (A) List all critical numbers of f. If there are n

Question

Suppose that

f(x)=3/(x^2-16).

(A) List all critical numbers of f. If there are no critical numbers, enter 'NONE'.
Critical numbers =

(B) Use interval notation to indicate where f(x) is increasing.

Note: Use 'INF' for ?, '-INF' for ??, and use 'U' for the union symbol.
Increasing:

(C) Use interval notation to indicate where f(x) is decreasing.

Decreasing:

(D)List the x-coordinates of all local maxima of f. If there are no local maxima, enter 'NONE'.

x values of local maxima =

(E) List the x-coordinates of all local minima of f. If there are no local minima, enter 'NONE'.

x values of local minima =

(F) Use interval notation to indicate where f(x) is concave up.

Concave up:

(G) Use interval notation to indicate where f(x) is concave down.

Concave down:

(H) List the x values all inflection points of f. If there are no inflection points, enter 'NONE'.

Inflection points =

(I) List all horizontal asymptotes of f. If there are no horizontal asymptotes, enter 'NONE'.

Horizontal asymptotes y=

(J) List all vertical asymptotes of f. If there are no vertical asymptotes, enter 'NONE'.

Vertical asymptotes x=

(K) Use all of the preceding information to

Explanation / Answer

f(x) = 3/(x^2-16)
f'(x) = -3 * 2x / (x^2 - 16)^2
f'(x) = -6x / (x^2 - 16)^2

(a)
For critical numbers, f'(x) = 0
-6x / (x^2 - 16)^2 = 0
x = 0 is the critical number

(b)
For increasing, f'(x) > 0
-6x / (x^2 - 16)^2 > 0
6x / (x^2 - 16)^2 < 0
x < 0
x = (-INF, 0)

(c)
For decreasing, f'(x) < 0
-6x / (x^2 - 16)^2 < 0
6x / (x^2 - 16)^2 > 0
x > 0
x = (0, INF)

(d)
f'(x) = -6x / (x^2 - 16)^2
f''(x) = -6 / (x^2 - 16)^2 + 12x * 2x / (x^2 - 16)^3
f''(x) = -6 / (x^2 - 16)^2 + 24x^2 / (x^2 - 16)^3
at critical number, x = 0, we have
f''(0) = -6 / (0 - 16)^2 + 24 * 0 / (0 - 16)^3
f''(0) = -6 / 256 = -3/128 < 0
Hence we have local maxima at x = 0

(e)
We do not have any local minima

(f)
f''(x) = -6 / (x^2 - 16)^2 + 24x^2 / (x^2 - 16)^3
For concave up, f''(x) > 0
f''(x) = [-6(x^2 - 16) + 24x^2] / (x^2 - 16)^3 > 0
f''(x) = [-6x^2 + 96 + 24x^2] / (x^2 - 16)^3 > 0
f''(x) = [18x^2 + 96] / (x - 4)^3 (x + 4)^3 > 0
x > 4, x < -4
x = (-INF, -4) U (4, INF)

(g)
For concave down, it will be then
x = (-4, 4)

(h)
For inflection points, f''(x) = 0
which is not possible here
Hence NONE

(i)
If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is the line y = 0.

(j)
vertical asymptote exist at denominator = 0
x^2 - 16 = 0
x = 4, -4

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.