Mathematics, 03.08.2020 14:01 dusjxjxxb2964

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x)

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Answer from: YODIIZ6590

2.25

Step-by-step explanation:

The computation of the number c that satisfied is shown below:

Given that

$f(x) = \sqrt{x}$

Interval = (0,9)

According to the Rolle's mean value theorem,

If f(x) is continuous in {a,b) and it is distinct also

And, f(a) ≠ f(b) so its existance should be at least one value

i.e

$f^i(c) = \frac{f(b) - f(a)}{b -a }$

After this,

$f(x) = \sqrt{x} \\\\ f^i(x) = \frac{1}{2}x ^{\frac{1}{2} - 1} \\\\ = \frac{1}{2}x ^{\frac{-1}{2}$

$f^i(x) = \frac{1}{{2}\sqrt{x} } = f^i(c) = \frac{1}{{2}\sqrt{c} } \\\\\a = 0, f (a) = f(o) = \sqrt{0} = 0 \\\\\ b = 9 , f (b) = f(a) = \sqrt{9} = 3\\$

After this,

Put the values of a and b to the above equation

$f^i(c) = \frac{f(b) - f(a)}{b - a} \\\\ \frac{1}{{2}\sqrt{c} } = \frac{3 -0}{9-0} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{3}{9} \\\\ \frac{1}{\sqrt[2]{c} } = \frac{1}{3} \\\\ \sqrt[2]{c} = 3\\\\\sqrt{c} = \frac{3}{2} \\\\ c = \frac{9}{4}$

= 2.25

Answer from: Quest

part d

step-by-step explanation:

theresa talked her parents into getting sd cards for her phone and her brother’s phone. inserting an sd card into a phone gives it more storage. they both get 8-gigabyte sd cards. how many bits of storage come with each sd card? type your answer in both scientific and standard notation.

Answer from: Quest

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: