Consider the following function. F(x) = 1 + 7/x - 4/x^2 Find the vertical asympt
ID: 2880319 • Letter: C
Question
Consider the following function. F(x) = 1 + 7/x - 4/x^2 Find the vertical asymptote(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = Find the horizontal asymptote(s). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Find the interval where the function is increasing. (Enter your answer using interval notation.) Find the interval where the function is decreasing. (Enter your answer using interval notation.) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.) local maximum value local minimum value Find the interval where the function is concave up. (Enter your answer using interval notation.) Find the interval where the function is concave down. (Enter your answer using interval notation.) Find the inflection point. (x y) =Explanation / Answer
given f(x)=1+(7/x)-(4/x2)
=>f(x)=((x2+7x-4)/x2)
domain is (-,0)U(0,)
a) for vertical asymptotes denominator =0
=>x2=0
=> x=0 is vertical asymptote
horizontal asymptote y=limx->f(x)
horizontal asymptote is y=limx->(1+(7/x)-(4/x2))
horizontal asymptote is y=(1+0-0)
horizontal asymptote is y=1
b)f(x)=1+(7/x)-(4/x2)
differentiate with respect to x
f '(x)=0-(7/x2)+(8/x3)
f '(x)=-(7/x2)+(8/x3)
f '(x)=((-7x+8)/x3)
function is increasing when f '(x)>0
=>((-7x+8)/x3) >0
=> interval is (0,8/7)
function is decreasing when f '(x)<0
=>((-7x+8)/x3) <0
=> interval is (-,0)U(8/7,)
c) for extremum f '(x)=0
=>((-7x+8)/x3)=0
=> x =8/7
at x =8/7, f '(x) changes from positive to negative.
so local maximum at x =8/7
no local minimum
local maximum value = f(8/7)
local maximum value = 1+(7/(8/7))-(4/(8/7)2)
local maximum value = 65/16
d)f '(x)=-(7/x2)+(8/x3)
differentiate with respect to x
f ''(x)=(14/x3)-(24/x4)
f ''(x)=(14x-24)/x4
concave up when f "(x)>0
(14x-24)/x4>0
x>12/7
concave up in (12/7,)
concave down when f "(x)<0
(14x-24)/x4<0
x><12/7
concave down in (-,0)U(0,12/7)
for inflection point f "(x)=0
=> x=12/7
f(12/7)=67/18
inflection point is (12/7,67/18)
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