f-x-x-7-2x-1-x-2-x-2-7x-4-x-lt-1-f-1-

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

f (x) = {\begin{cases} x^{7} + 2 x + 1; x ⩾ 2 \\ x^{2} + 7 x + 4; x < 1 \end{cases}

f^{^{'}} (1) = ?

Answered by MM42 last updated on 25/Jul/23

f (1), n o t a v a i l a b l e s o

n o t a v a i l a b l e f^{'} .

Answered by dimentri last updated on 25/Jul/23

f^{'} (1) = \lim_{x \to 1^{+}} f^{'} (x) = \lim_{x \to 1^{-}} f^{'} (x)

= \lim_{x \to 1^{+}} (7 x^{6} + 2) = 9

= \lim_{x \to 1^{-}} (2 x + 7) = 9

\begin{matrix} \underset{⏟}{W} \end{matrix}

Commented by MM42 last updated on 25/Jul/23

f^{'} (a) = {l i m}_{x \to a} \frac{f (x) - f (a)}{x - a}

f_{+}^{'} (a) = {l i m}_{x \to a^{+}} \frac{f (x) - f (a)}{x - a}

f_{-}^{'} (a) = {l i m}_{x \to a^{-}} \frac{f (x) - f (a)}{x - a}

t h e r e f o r e, t h e r e m o s t a l w a y s b e “ f {(a)}^{″}

f o r x ⩽ 2 \to f^{'} (x) = 7 x^{6} + 2 \Rightarrow f_{+}^{'} (2) = 7 \times 64 + 2 = 450 = f^{'} (2)

i f x < 1 \to f^{'} (x) = 2 x + 7 . b u t n o t

e x i s t f^{'} (1)

Commented by mathlove last updated on 26/Jul/23

w h a t i s t h e v a l u e

f_{-}^{^{'}} (1) = ?

Commented by MM42 last updated on 26/Jul/23

n o t e x i s t . b e c a u s e n o t e x i s t f (1)

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