Suppose that f(x)=3x^6-4x^5. (A) Find all critical numbers of f. If there are no

Question

Suppose that

f(x)=3x^6-4x^5.

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

x values of local maxima =

(E) Find the x-coordinates of all local minima of f. Note: 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 of all inflection points of f. If there are no inflection points, enter 'NONE'.

x values of inflection points =

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

Horizontal asymptotes y=

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

Vertical asymptotes x=

(K) Use all of the preceding information to sketch a graph of f. When you're finished, enter a "1" in the box below.

Graph Complete:

Explanation / Answer

f(x) = 3x^6 - 4x^5
f'(x) = 18x^5 - 20x^4
f'(x) = 2x^4(9x - 10)

(a)
For critical numbers, f'(x) = 0
2x^4(9x - 10) = 0
x = 0, 10/9 are the critical numbers

(b)
For increasing, f'(x) > 0
2x^4(9x - 10) > 0
x < 0, x > 10/9
x = (-INF, 0) U (10/9, INF)

(c)
For decreasing, f'(x) < 0
x = (0, 10/9)

(d)
f'(x) = 18x^5 - 20x^4
f''(x) = 90x^4 - 80x^3
f''(x) = 10x^3(9x - 8)
at critical number, x = 0, we have
f''(0) =0
at critical number, x = 10/9, we have
f''(x) = 10 * 10/9^3(9 * 10/9 - 8) > 0
Hence x = 0 is inflection point and x = 10/9 is local minima
So we have NONE local maxima

(e)
x = 10/9 is local minima

(f)
f''(x) = 10x^3(9x - 8)
For concave up, f''(x) > 0
f''(x) = 10x^3(9x - 8) > 0
x > 0, x < 8/9
x = (-INF, 0) U (8/9, INF)

(g)
For concave down, it will be then
x = (0, 8/9)

(h)
For inflection points, f''(x) = 0
10x^3(9x - 8) = 0
x = 0, 8/9 are inflection points

(i)
If the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote.

(j)
vertical asymptote exist at denominator = 0
x = -infinity

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