Differentiable function proof given ##f''(c) = 1##

[member 731016]

Homework Statement

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Relevant Equations

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For this problem,

I'm confused by the implication from the antecedent $0 < |x - c| < \delta$ to the consequent. Should the consequent not be $|f''(x) - f''(c)| < \frac{1}{2}$ where $\epsilon = \frac{1}{2}$ (Since we are applying the definition of a limit for the first derivative curve)?

$|f''(x) - f''(c)| < \frac{1}{2}$

$\leftrightarrow |f''(x) - \frac{f'(x) - f'(c)}{x - c}| < \frac{1}{2}$

$\leftrightarrow |f''(x) - 1| < \frac{1}{2}$

I don't understand why they don't have $f''(x)$ in their expression. Does someone please know why?

Thanks!

 

[fresh_42]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 346372
I'm confused by the implication from the antecedent $0 < |x - c| < \delta$ to the consequent. Should the consequent not be $|f''(x) - f''(c)| < \frac{1}{2}$ where $\epsilon = \frac{1}{2}$ (Since we are applying the definition of a limit for the first derivative curve)?

Why? We already have $f''(c)=1.$ The statement is just a reformulation of this fact by using the definition of a derivative, choosing $\varepsilon =0.5$ and the definition of a limit with this given margin $\varepsilon .$ The proof goes:

\begin{align*}
1&=f''(c)\stackrel{\text{short for}}{=}\left. \dfrac{d}{dx} \right|_{x=c}f'(x)\stackrel{\text{def. of derivative of }f'}{=}\lim_{x \to c}\dfrac{f'(x)-f'(c)}{x-c}\\[12pt]
0&=\left(\lim_{x \to c}\dfrac{f'(x)-f'(c)}{x-c}\right)-1=\lim_{x \to c}\left(\dfrac{f'(x)-f'(c)}{x-c} - 1\right)\\[12pt]
&\left|\dfrac{f'(x)-f'(c)}{x-c}-1\right| \stackrel{\text{def. of limit at }x=c}{<} \varepsilon :=\dfrac{1}{2} \text{ for all } |x-c|<\delta\\[12pt]
&\substack{\text{definition of} \\ \text{the absolute values}}\quad\dfrac{1}{2}=1-\varepsilon <\dfrac{f'(x)-f'(c)}{x-c}<1+\varepsilon =\dfrac{3}{2} \text{ for all } c-\delta <x<c+\delta
\end{align*}