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2 years ago

12

What are the domain and range of the function below?

2 answers:

irina1246 [14]2 years ago

8 0

I think the answer is

domain : (- infinity , infinity )

range : [ -2 , infinity )

i am sorry if my answer was wrong

kompoz [17]2 years ago

3 0

Answer:

C. domain is - infinity, infinity

range is -2, infinity

Step-by-step explanation:

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Mr. King divides his entire class into 8 stations. He assigns 4 students to each station. How many students are in the class?

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2 years ago

Read 2 more answers

Use the intermediate value theorem to find the value of c such that f(c) = M. f(x) = x^2 - x + 1 text( on ) [1,8]; M = 21 c =

DanielleElmas [232]

Answer:

$c = 5$

Step-by-step explanation:

Given

$f(c) = M$

$f(x) = x^2 - x + 1$

$Interval: [1,8]$

$M = 21$

Required

Find c using Intermediate Value theorem

First, check if the value of M is within the given range:

$f(x) = x^2 - x + 1$

$f(1) = 1^2 - 1 + 1$

$f(1) = 1$

$f(x) = 8^2 - 8 + 1$

$f(x) = 57$

$1 \le M \le 57$

$1 \le 21 \le 57$

M is within range.

Solving further:

We have:

$f(c) = f(x) = M$

$f(x) = 21$

Substitute 21 for f(x) in $f(x) = x^2 - x + 1$

$21 = x^2 - x + 1$

Express as quadratic function

$x^2 - x + 1 - 21 = 0$

$x^2 - x - 20 = 0$

Expand

$x^2 + 4x - 5x - 20$

$x(x+4)-5(x+4)=0$

$(x - 5)(x+4) = 0$

$x - 5 = 0$ or $x + 4= 0$

$x = 5$ or $x = -4$

The value of $x = -4$ is outside the $Interval: [1,8]$

So:

$x = 5$

$f(c) = f(x) = M$

$f(c) = f(5) = 21$

By comparison:

$c = 5$

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