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I have to find the types of roots (i.e real or complex) of the equation $$ 11^x + 13^x+ 17^x -19^x = 0 \dots (1) $$

If $$ f(x) = 11^x + 13^x+ 17^x -19^x = 0 $$ , then obviously $ f'(x)= 0 $ has a 0 solution, and indeed every derivative of $f(x)$ has a 0 solution.

In this context a question arises in my mind : if all the conditions of Rolle's Theorem are satisfied for a function $g(x) $ in $[a,b]$, and in addition if $g'(c)=0$ ,then is it necessary that $c$ lies between $a$ and $b$ ?

If it's true, then we can conclude that $f(x)=0 $ has more than 2 real roots right?

Any insight ? Thank you.

Jean Marie
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hiren_garai
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1 Answers1

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if all the conditions of Rolle's Theorem are satisfied for a function $g(x) $ in $[a,b]$, and in addition if $g'(c)=0$ ,then is it necessary that $c$ lies between $a$ and $b$ ?

No, not at all. Consider for example, $g(x)=\sin(x)$ with $a=0$, $b=\pi$, and $c=\frac32\pi$.

Or (if you mean "strictly between $a$ and $b$"), consider $g(x)=\sin^2(x)$ with $a=0$ and $b=c=\pi$.

hmakholm left over Monica
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  • Thanks for the counterexamples ! @HenningMakholm. – hiren_garai Sep 02 '17 at 17:07
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    This is a good example, but the larger point is that Rolle's Thm. assumes nothing about the function $f (x)$ outside the closed interval $[a,b]$, and cannot conclude anything about $f(x)$ outside the interval. – hardmath Sep 02 '17 at 17:10