Let $f$ be a quadratic function. Prove the Typical Value Theorem uses to $f$ on any closed interval $$ and that the "$c$" guaranteed by the Mean Value Theorem is the midallude of the interval.

How do I go about addressing this? I understand also just how the Average Value Theorem functions yet I don"t recognize exactly how it have the right to prove that the c is the midsuggest of the interval. Thanks in advanced.


*

Notice that this especially claims "quadratic function". This isn"t true for just any type of function, f. Write $f(x)= px^2+ qx+ r$. The rest of this is just computation. At x= a that is $pa^2+ qa+ r$. At x= b it is $pb^2+ qb+ r$. $fracf(b)- f(a)b- a= fracpb^2- qb+ r- pa^2- qa+ rb- a= fracp(b^2- a^2)+ q(b- a)b- a= fracp(b+a)(b-a)+ q(b-a)b-a= p(b+a)+ q$.

You are watching: Find the value of c guaranteed by the mean value theorem

On the various other hand $f"= 2px+ q$. The "intend worth theorem" claims that there exit a allude, c, such that $f"(c)= fracf(b)-f(a)b- a$. Here that implies that $2pc+ q= p(b+a)+ q$. From that $2pc= p(b+a)$ so that $c= fracb+a2$


Share
Cite
Follow
edited Oct 31 "18 at 0:49
answered Oct 31 "18 at 0:39
*

user247327user247327
17.6k22 gold badges1010 silver badges2020 bronze badges
$endgroup$
Add a comment |
0
$egingroup$
In order to use MVT, we recognize that $f$ need to be consistent on the closed interval $$ and differentiable on $(a, b)$. Then MVT guarantees that tbelow exists$$f ^ prime ( c ) = frac f ( b ) - f ( a ) b - a $$You desire to present that $c = fraca+b2$ as this is the midallude between $a, b$.

We can verify this with algebraic simplification and also substitution. Let $f(x) = px^2+qx+z$. Then $f"(x) = 2px + q$ such that $a eq 0$.

Plug in $a, b$ in for $x$ in $f(x) = px^2+qx+z$ and then plug that expression into $frac f ( b ) - f ( a ) b - a $.

You have to then get $p(b+a) + q$

Then plug $c$ in for $x$ in $f"(x)$ and you need to gain $pc + q$. Keep in mind that these two worths are equivalent according to MVT. Or if you substitute $fraca+bc$ for $c$ in $f"(x)$ then you will certainly view that the expressions are tantamount.


Share
Cite
Follow
answered Oct 31 "18 at 0:43
*

moloculemolocule
16277 bronze badges
$endgroup$
Add a comment |

Your Answer


Thanks for contributing an answer to slrfc.orgematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid

Asking for aid, clarification, or responding to other answers.Making statements based on opinion; earlier them up through references or personal experience.

Use slrfc.orgJax to format equations. slrfc.orgJax recommendation.

To learn more, see our tips on writing excellent answers.

See more: Careers For People Who Get Bored Easily (With Salaries), 6 Careers For Those Who Get Bored Easily


Draft saved
Draft discarded

Sign up or log in


Sign up utilizing Google
Sign up using Facebook
Sign up using Email and Password
Submit

Message as a guest


Name
Email Required, but never shown


Article as a guest


Name
Email

Required, but never shown


Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our regards to business, privacy policy and cookie plan


Not the answer you're looking for? Browse other inquiries tagged calculus or ask your own question.


Featured on Meta
Related
1
Prove, using the mean worth theorem, that $x+1 lt e^x lt xe^x+1$ for $x gt 0$
0
Finding all numbers guaranteed by the Median Value Theorem?
0
Proof utilizing the suppose value theorem
11
Why does the Average Value Theorem call for a closed interval for continuity and an open up interval for differentiability?
12
Difficulties in stating intend value theorem for features which are not continuous on a closed interval.
0
For which attributes is the ‘c’ from the expect value theorem the midallude of the interval?
1
How to usage the Mean-Value Theorem on a function constant over an open interval.
1
Prove using the Typical Value Theorem
0
Prove Using L'Hopital's Rule And Typical Value Theorem.
Hot Netoccupational Questions more hot inquiries

Question feed
Subscribe to RSS
Inquiry feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader.


*

slrfc.orgematics
Company
Stack Exchange Network-related
website architecture / logo © 2021 Stack Exadjust Inc; user contributions licensed under cc by-sa. rev2021.9.16.40224


slrfc.orgematics Stack Exreadjust works best with JavaScript enabled
*

Your privacy

By clicking “Accept all cookies”, you agree Stack Exadjust have the right to store cookies on your gadget and also discshed indevelopment in accordance with our Cookie Policy.