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Sagot :
Answer:
See below
Step-by-step explanation:
I assume the function is [tex]f(x)=1+\frac{5}{x}-\frac{4}{x^2}[/tex]
A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, [tex]x=0[/tex] is the only vertical asymptote.
B) Set the first derivative equal to 0 and solve:
[tex]f(x)=1+\frac{5}{x}-\frac{4}{x^2}[/tex]
[tex]f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}[/tex]
[tex]0=-\frac{5}{x^2}+\frac{8}{x^3}[/tex]
[tex]0=-5x+8[/tex]
[tex]5x=8[/tex]
[tex]x=\frac{8}{5}[/tex]
Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:
[tex]f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}[/tex]
[tex]f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3[/tex]
Therefore, the function increases on the interval [tex](0,\frac{8}{5})[/tex] and decreases on the interval [tex](-\infty,0),(\frac{8}{5},\infty)[/tex].
C) Since we determined that the slope is 0 when [tex]x=\frac{8}{5}[/tex] from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, [tex]f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}[/tex], meaning there's an extreme at the point [tex](\frac{8}{5},\frac{41}{16})[/tex], but is it a maximum or minimum? To answer that, we will plug in [tex]x=\frac{8}{5}[/tex] into the second derivative which is [tex]f''(x)=\frac{10}{x^3}-\frac{24}{x^4}[/tex]. If [tex]f''(x)>0[/tex], then it's a minimum. If [tex]f''(x)<0[/tex], then it's a maximum. If [tex]f''(x)=0[/tex], the test fails. So, [tex]f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}<0[/tex], which means [tex](\frac{8}{5},\frac{41}{16})[/tex] is a local maximum.
D) Now set the second derivative equal to 0 and solve:
[tex]f''(x)=\frac{10}{x^3}-\frac{24}{x^4}[/tex]
[tex]0=\frac{10}{x^3}-\frac{24}{x^4}[/tex]
[tex]0=10x-24[/tex]
[tex]-10x=-24[/tex]
[tex]x=\frac{24}{10}[/tex]
[tex]x=\frac{12}{5}[/tex]
We then test where [tex]f''(x)[/tex] is negative or positive by plugging in test values. I will use -1 and 3 to test this:
[tex]f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34<0[/tex], so the function is concave down on the interval [tex](-\infty,0)\cup(0,\frac{12}{5})[/tex]
[tex]f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0[/tex], so the function is concave up on the interval [tex](\frac{12}{5},\infty)[/tex]
The inflection point is where concavity changes, which can be determined by plugging in [tex]x=\frac{12}{5}[/tex] into the original function, which would be [tex]f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}[/tex], or [tex](\frac{12}{5},\frac{43}{18})[/tex].
E) See attached graph
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