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Exercise 7.3 · Q5

Q.Show that the value in the conclusion of the mean value theorem for

(i) f(x)=1xf(x)=\dfrac1x on a closed interval of positive numbers [a,b][a,b] is ab\sqrt{ab}
(ii) f(x)=Ax2+Bx+Cf(x)=Ax^2+Bx+C on any interval [a,b][a,b] is a+b2\dfrac{a+b}{2}.
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Both parts are proved the same way: write the LMVT equation symbolically for the general function, simplify the right side algebraically, and solve for cc.

Step 1 (i). f(x)=1xf(x)=\dfrac1x on [a,b][a,b], a,b>0a,b>0.

f′(x)=−1x2f'(x)=-\dfrac{1}{x^2}.

f(b)−f(a)b−a=1b−1ab−a=a−babb−a=−(b−a)ab(b−a)=−1ab.\frac{f(b)-f(a)}{b-a}=\frac{\tfrac1b-\tfrac1a}{b-a}=\frac{\tfrac{a-b}{ab}}{b-a}=\frac{-(b-a)}{ab(b-a)}=-\frac{1}{ab}.

Set f′(c)=−1c2f'(c)=-\dfrac{1}{c^2} equal to this: −1c2=−1ab⇒c2=ab⇒c=ab-\dfrac{1}{c^2}=-\dfrac{1}{ab}\Rightarrow c^2=ab\Rightarrow c=\sqrt{ab} (taking the positive root, since c∈(a,b)c\in(a,b) with a,b>0a,b>0).

Step 2 (ii). f(x)=Ax2+Bx+Cf(x)=Ax^2+Bx+C on [a,b][a,b].

f′(x)=2Ax+Bf'(x)=2Ax+B.

f(b)−f(a)b−a=A(b2−a2)+B(b−a)b−a=A(a+b)+B.\frac{f(b)-f(a)}{b-a}=\frac{A(b^2-a^2)+B(b-a)}{b-a}=A(a+b)+B. …

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