Divide and simplify.

Whats is the zero of function ?<br>A -3<br>B-3/2<br>C-2/3<br>D-2

Brrunno [24]

Answer:

D. -2

Step-by-step explanation:

The zero of a function is always along the x-axis and in this case that's -2.

Hope this helps :)

4 0

3 years ago

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

The classes at the middle school want to raise money.

Gnesinka [82]

Answer:

This is a multiple step question. Before you can answer it you must find what it is asking, which is the highest rate for making money, or the greatest rate of change. To compare the rate of change between each grade, you must first have them in common units, which can be done by finding how much money was made by each grade in one hour. To find this, divide the number of dollars made by the number of hours it was made in to get how much was made in one hour. Doing this shows that the sixth grade made 34 dollars an hour (170 divided by 5), seventh made 28 dollars an hour (112/4), and the eighth grade made 32 dollars an hour (192/6). Then compare the numbers to see that the greatest amount earned in one hour was 34 dollars, by the sixth grade, which then becomes your answer: The sixth grade had the highest rate for raising money.

6 0

3 years ago

Determine whether the function f(x)=x-3x^3 is even, odd, or neither.

vredina [299]

Answer:

Odd

Step-by-step explanation:

Looking at f(x)=x-3x^3, we see that all powers of x are odd: x^1 and x^3.

Thus, this function is ODD.

5 0

3 years ago

Help??? Answer and explain

bezimeni [28]

The relation is not a function because the input numbers have a number that is repeated twice. That number is 9.

6 0

4 years ago