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 $c= fracb+a2$

edited Oct 31 "18 at 0:49
answered Oct 31 "18 at 0:39

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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.

answered Oct 31 "18 at 0:43

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