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Question Number 195170 by mathlove last updated on 25/Jul/23

$$f\left(x\right)=\{\begin{array}{l}{x}^7+2x+1;x⩾2\\ x^2+7x+4;x<1\end{array}$$$$f^{{}^{\prime }}\left(1\right)=?$$

Answered by MM42 last updated on 25/Jul/23

$$f\left(1\right),notavailableso$$$$notavailable{f}^{\prime }.$$

Answered by dimentri last updated on 25/Jul/23

$$f{}^{\prime }\left(1\right)=\underset{x\to 1^{+}}{\lim }f{}^{\prime }\left(x\right)=\underset{x\to 1^{-}}{\lim }f{}^{\prime }\left(x\right)$$$$=\underset{x\to 1^{+}}{\lim }(7x^6+2)=9$$$$=\underset{x\to 1^{-}}{\lim }(2x+7)=9$$$$\overline{)\begin{array}{c}\underbrace{\mathcal{W}}\end{array}}$$

Commented by MM42 last updated on 25/Jul/23

$$f^{\prime }\left(a\right)={lim}_{x\to a}\frac{f\left(x\right)-f\left(a\right)}{x-a}$$$$f_{+}^{\prime }\left(a\right)={lim}_{x\to a^{+}}\frac{f\left(x\right)-f\left(a\right)}{x-a}$$$$f_{-}^{\prime }\left(a\right)={lim}_{x\to a^{-}}\frac{f\left(x\right)-f\left(a\right)}{x-a}$$$$therefore,theremostalwaysbe‘‘f{\left(a\right)}^{″}$$$$$$$$forx⩽2\to f{}^{\prime }\left(x\right)=7x^6+2\Rightarrow f_{+}^{\prime }\left(2\right)=7\times 64+2=450=f^{\prime }\left(2\right)$$$$ifx<1\to f^{\prime }\left(x\right)=2x+7.butnot$$$$exist{f}^{\prime }\left(1\right)$$

Commented by mathlove last updated on 26/Jul/23

$$whatisthevalue$$$$f_{-}^{{}^{\prime }}\left(1\right)=?$$

Commented by MM42 last updated on 26/Jul/23

$$notexist.becausenotexistf\left(1\right)$$

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