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Lisa [10]
3 years ago
13

Can anyone answer this?

Mathematics
2 answers:
il63 [147K]3 years ago
6 0

Answer:

D

Step-by-step explanation:

3241004551 [841]3 years ago
3 0

Answer:

yes d

Step-by-step explanation:

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Consider the following function.
Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is f(x)=1+\frac{5}{x}-\frac{4}{x^2}

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, x=0 is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

f(x)=1+\frac{5}{x}-\frac{4}{x^2}

f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}

0=-\frac{5}{x^2}+\frac{8}{x^3}

0=-5x+8

5x=8

x=\frac{8}{5}

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}

f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3

Therefore, the function increases on the interval (0,\frac{8}{5}) and decreases on the interval (-\infty,0),(\frac{8}{5},\infty).

C) Since we determined that the slope is 0 when x=\frac{8}{5} from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}, meaning there's an extreme at the point (\frac{8}{5},\frac{41}{16}), but is it a maximum or minimum? To answer that, we will plug in x=\frac{8}{5} into the second derivative which is f''(x)=\frac{10}{x^3}-\frac{24}{x^4}. If f''(x)>0, then it's a minimum. If f''(x), then it's a maximum. If f''(x)=0, the test fails. So, f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}, which means (\frac{8}{5},\frac{41}{16}) is a local maximum.

D) Now set the second derivative equal to 0 and solve:

f''(x)=\frac{10}{x^3}-\frac{24}{x^4}

0=\frac{10}{x^3}-\frac{24}{x^4}

0=10x-24

-10x=-24

x=\frac{24}{10}

x=\frac{12}{5}

We then test where f''(x) is negative or positive by plugging in test values. I will use -1 and 3 to test this:

f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34, so the function is concave down on the interval (-\infty,0)\cup(0,\frac{12}{5})

f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0, so the function is concave up on the interval (\frac{12}{5},\infty)

The inflection point is where concavity changes, which can be determined by plugging in x=\frac{12}{5} into the original function, which would be f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}, or (\frac{12}{5},\frac{43}{18}).

E) See attached graph

5 0
3 years ago
 PLEASE HELP 15 PTS  <br>HELP ME WITH TRUE OR FALSE MATH 
Sever21 [200]
  • 1st on the top right hand corner is True
  • 2nd on the second row to the right is True
  • 3rd on the first row, second column is False
  • 4th on the second column second row is False
  • 5th on the third column, first row is True
  • 6th on the third column, second row is True
8 0
3 years ago
If 60% of the students in the classroom have a pet at home. If 15 students have a pet, what is the total number of students in t
torisob [31]
There are 25 students in the classroom
3 0
3 years ago
For the function f(x)=5x^3-6, find f^-1
max2010maxim [7]

Answer:

f^{-1} means the inverse function

f^{-1} = \sqrt[3]{(x + 6)/5}

Step-by-step explanation:

f^{-1} means the inverse function

to find an inverse function, flip the x and y variables and solve for y

f(x) = 5x^3 - 6

y = 5x^3 - 6

x = 5y^3 - 6

x + 6 = 5y^3

(x + 6)/5 = y^3

\sqrt[3]{(x + 6)/5} = y

5 0
3 years ago
Which scenarios represent one quantity changing at a constant rate per unit interval relative to the other quantity?
irinina [24]

Answer:

Graphs come in all sorts of shapes and sizes. In algebra, there are 3 basic types of graphs you'll see most often: linear, quadratic, and exponential. Check out this tutorial and learn how to determine is a graph represents a linear, quadratic, or exponential function!

Step-by-step explanation:

thats all

<h2 />
5 0
3 years ago
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