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\$

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edited Oct 31 "18 at 0:49
answered Oct 31 "18 at 0:39 user247327user247327
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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.

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answered Oct 31 "18 at 0:43 moloculemolocule
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