ANSWER ALL THE QUESTION(S) PLEASE Find the inflection point(s), if any, of the f

Question

ANSWER ALL THE QUESTION(S) PLEASE

Find the inflection point(s), if any, of the function. (If an answer does not exist, enter DNE.)

g(x) = 2x3 3x2 + 6x 10

Find the relative extrema, if any, of the function. Use the Second Derivative Test if applicable. (If an answer does not exist, enter DNE.)

g(x) = x3 9x Relative Max:?? Relative min:?? Show answer in (x,y) form.

Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph. If an answer does not exist, enter DNE.)

f(x)= 2x/(x^2-3x-10) Find the 2 vertical asymptotes, larger and smaller values.

Explanation / Answer

ANSWER---

    Now—

f(1/2) = 2/8 - 3/4 + 6/2 - 10
= (2 - 6 + 24 - 40) / 8
= -20/8
= -5/2

The inflection point is (1/2,-5/2).

0 = = 3x^2-9


The second derivative test for local extrema: If f '(x) = 0 and f ''(x) > 0, then f(x) has a local minimum at x0. If f '(x) = 0 and f ''(x) < 0, then f(x) has a local maximum at x.

f''(x) = 6x
f''(-(sqr root 3))=-6(sqr root 3) is a local maximum
f''((sqr root 3)) =6(sqr root 3)   is a local minimum

3) f(x)= 2x/(x^2-3x-10)

The vertical asymptotes come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve.

(x^2-3x-10)=0

Since the degree is greater in the denominator than in the numerator, the y-values will be dragged down to the x-axis, and the horizontal asymptote is therefore "y = 0". Since I have found a horizontal asymptote, I don't have to look for a slant asymptote.

So the horizontal asymptote is at y=0 (x axis)

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