g ( x ) = x ∣ x ∣ g ( x ) = x | x | g ( x ) = x ∣ x ∣ h ( x ) = ∣ x − 1 ∣ 3 h ( x ) = | x - 1 | ^{3} h ( x ) = ∣ x − 1 ∣ 3 g ( x ) g ( x ) g ( x ) h ( x ) h ( x ) h ( x ) f ( x ) f ( x ) f ( x ) g ′ ( 0 ) = 0 g' ( 0 ) = 0 g ′ ( 0 ) = 0 h ′ ( 1 ) = 0 h' ( 1 ) = 0 h ′ ( 1 ) = 0 x > 0 x > 0 x > 0 g ( x ) = x 2 g ( x ) = x ^{2} g ( x ) = x 2 x < 1 x < 1 x < 1 h ( x ) = − ( x − 1 ) 3 h ( x ) = - ( x - 1 ) ^{3} h ( x ) = − ( x − 1 ) 3 f ′ ( 0 ) = 0 + h ′ ( 0 ) f' ( 0 ) = 0 + h' ( 0 ) f ′ ( 0 ) = 0 + h ′ ( 0 ) = − 3 ( 0 − 1 ) 2 = − 3 = - 3 ( 0 - 1 ) ^{2} = - 3 = − 3 ( 0 − 1 ) 2 = − 3 f ′ ( 1 ) = g ′ ( 1 ) + 0 = 2 ⋅ 1 = 2 f' ( 1 ) = g' ( 1 ) + 0 = 2 \cdot 1 = 2 f ′ ( 1 ) = g ′ ( 1 ) + 0 = 2 ⋅ 1 = 2 ∴ \therefore ∴ f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1 f' ( 0 ) + f' ( 1 ) = - 3 + 2 = - 1 f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1 f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ 3 f ( x ) = x | x | + | x - 1 | ^{3} f ( x ) = x ∣ x ∣ + ∣ x − 1 ∣ 3 f ( 0 ) = 1 f ( 0 ) = 1 f ( 0 ) = 1 f ( 1 ) = 1 f ( 1 ) = 1 f ( 1 ) = 1 lim h → + 0 f ( 0 + h ) − f ( 0 ) h \displaystyle \lim\limits _{h \rightarrow + 0} \frac{f ( 0 + h ) - f ( 0 )}{h} h → + 0 lim h f ( 0 + h ) − f ( 0 ) = lim h → + 0 h ∣ h ∣ + ∣ h − 1 ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{h | h | + | h - 1 | ^{3} - 1}{h} = h → + 0 lim h h ∣ h ∣ + ∣ h − 1 ∣ 3 − 1 = lim h → + 0 h 2 − ( h − 1 ) 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{h ^{2} - ( h - 1 ) ^{3} - 1}{h} = h → + 0 lim h h 2 − ( h − 1 ) 3 − 1 = lim h → + 0 − h 3 + 4 h 2 − 3 h h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{- h ^{3} + 4 h ^{2} - 3 h}{h} = h → + 0 lim h − h 3 + 4 h 2 − 3 h = lim h → + 0 ( − h 2 + 4 h − 3 ) \displaystyle = \lim\limits _{h \rightarrow + 0} {} ( - h ^{2} + 4 h - 3 ) = h → + 0 lim ( − h 2 + 4 h − 3 ) = − 3 = - 3 = − 3 lim h → − 0 f ( 0 + h ) − f ( 0 ) h \displaystyle \lim\limits _{h \rightarrow - 0} {} \frac{f ( 0 + h ) - f ( 0 )}{h} h → − 0 lim h f ( 0 + h ) − f ( 0 ) = lim h → − 0 h ∣ h ∣ + ∣ h − 1 ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{h | h | + | h - 1 | ^{3} - 1}{h} = h → − 0 lim h h ∣ h ∣ + ∣ h − 1 ∣ 3 − 1 = lim h → − 0 − h 2 − ( h − 1 ) 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{- h ^{2} - ( h - 1 ) ^{3} - 1}{h} = h → − 0 lim h − h 2 − ( h − 1 ) 3 − 1 = lim h → − 0 − h 3 + 2 h 2 − 3 h h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{- h ^{3} + 2 h ^{2} - 3 h}{h} = h → − 0 lim h − h 3 + 2 h 2 − 3 h = lim h → − 0 ( − h 2 + 2 h − 3 ) \displaystyle = \lim\limits _{h \rightarrow - 0} {} ( - h ^{2} + 2 h - 3 ) = h → − 0 lim ( − h 2 + 2 h − 3 ) = − 3 = - 3 = − 3 ∴ \therefore ∴ f ′ ( 0 ) = − 3 f' ( 0 ) = - 3 f ′ ( 0 ) = − 3 lim h → + 0 f ( 1 + h ) − f ( 1 ) h \displaystyle \lim\limits _{h \rightarrow + 0} {} \frac{f ( 1 + h ) - f ( 1 )}{h} h → + 0 lim h f ( 1 + h ) − f ( 1 ) = lim h → + 0 ( 1 + h ) ∣ 1 + h ∣ + ∣ 1 + h − 1 ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{( 1 + h ) | 1 + h | + | 1 + h - 1 | ^{3} - 1}{h} = h → + 0 lim h ( 1 + h ) ∣1 + h ∣ + ∣1 + h − 1 ∣ 3 − 1 = lim h → + 0 ( 1 + h ) 2 + ∣ h ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{( 1 + h ) ^{2} + | h | ^{3} - 1}{h} = h → + 0 lim h ( 1 + h ) 2 + ∣ h ∣ 3 − 1 = lim h → + 0 h 3 + h 2 + 2 h h \displaystyle = \lim\limits _{h \rightarrow + 0} {} \frac{h ^{3} + h ^{2} + 2 h}{h} = h → + 0 lim h h 3 + h 2 + 2 h = lim h → + 0 ( h 2 + h + 2 ) \displaystyle = \lim\limits _{h \rightarrow + 0} {} ( h ^{2} + h + 2 ) = h → + 0 lim ( h 2 + h + 2 ) = 2 = 2 = 2 lim h → − 0 f ( 1 + h ) − f ( 1 ) h \displaystyle \lim\limits _{h \rightarrow - 0} {} \frac{f ( 1 + h ) - f ( 1 )}{h} h → − 0 lim h f ( 1 + h ) − f ( 1 ) = lim h → − 0 ( 1 + h ) ∣ 1 + h ∣ + ∣ 1 + h − 1 ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{( 1 + h ) | 1 + h | + | 1 + h - 1 | ^{3} - 1}{h} = h → − 0 lim h ( 1 + h ) ∣1 + h ∣ + ∣1 + h − 1 ∣ 3 − 1 = lim h → − 0 ( 1 + h ) 2 + ∣ h ∣ 3 − 1 h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{( 1 + h ) ^{2} + | h | ^{3} - 1}{h} = h → − 0 lim h ( 1 + h ) 2 + ∣ h ∣ 3 − 1 = lim h → − 0 − h 3 + h 2 + 2 h h \displaystyle = \lim\limits _{h \rightarrow - 0} {} \frac{- h ^{3} + h ^{2} + 2 h}{h} = h → − 0 lim h − h 3 + h 2 + 2 h = lim h → − 0 ( − h 2 + h + 2 ) \displaystyle = \lim\limits _{h \rightarrow - 0} {} ( - h ^{2} + h + 2 ) = h → − 0 lim ( − h 2 + h + 2 ) = 2 = 2 = 2 ∴ \therefore ∴ f ′ ( 1 ) = 2 f' ( 1 ) = 2 f ′ ( 1 ) = 2 ∴ \therefore ∴ f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1 f' ( 0 ) + f' ( 1 ) = - 3 + 2 = - 1 f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1 f ( x ) f ( x ) f ( x ) = { x 2 + ( x − 1 ) 3 ( x ≥ 1 ) x 2 − ( x − 1 ) 3 ( 0 ≤ x < 1 ) − x 2 − ( x − 1 ) 3 ( x < 0 ) \displaystyle = {\begin{cases} x ^{2} + ( x - 1 ) ^{3} & ( x \geq 1 ) \\ x ^{2} - ( x - 1 ) ^{3} & ( 0 \leq x < 1 ) \\ - x ^{2} - ( x - 1 ) ^{3} & ( x < 0 ) \end{cases}} = ⎩ ⎨ ⎧ x 2 + ( x − 1 ) 3 x 2 − ( x − 1 ) 3 − x 2 − ( x − 1 ) 3 ( x ≥ 1 ) ( 0 ≤ x < 1 ) ( x < 0 ) f ′ ( x ) = { 2 x + 3 ( x − 1 ) 2 ( x > 1 ) 2 x − 3 ( x − 1 ) 2 ( 0 < x < 1 ) − 2 x − 3 ( x − 1 ) 2 ( x < 0 ) \displaystyle f' ( x ) = {\begin{cases} 2 x + 3 ( x - 1 ) ^{2} & ( x > 1 ) \\ 2 x - 3 ( x - 1 ) ^{2} & ( 0 < x < 1 ) \\ - 2 x - 3 ( x - 1 ) ^{2} & ( x < 0 ) \end{cases}} f ′ ( x ) = ⎩ ⎨ ⎧ 2 x + 3 ( x − 1 ) 2 2 x − 3 ( x − 1 ) 2 − 2 x − 3 ( x − 1 ) 2 ( x > 1 ) ( 0 < x < 1 ) ( x < 0 ) lim x → + 0 f ′ ( x ) = 2 ⋅ 0 − 3 ( 0 − 1 ) 2 = − 3 \displaystyle \lim\limits _{x \rightarrow + 0} {} f' ( x ) = 2 \cdot 0 - 3 ( 0 - 1 ) ^{2} = - 3 x → + 0 lim f ′ ( x ) = 2 ⋅ 0 − 3 ( 0 − 1 ) 2 = − 3 lim x → − 0 f ′ ( x ) = − 2 ⋅ 0 − 3 ( 0 − 1 ) 2 = − 3 \displaystyle \lim\limits _{x \rightarrow - 0} {} f' ( x ) = - 2 \cdot 0 - 3 ( 0 - 1 ) ^{2} = - 3 x → − 0 lim f ′ ( x ) = − 2 ⋅ 0 − 3 ( 0 − 1 ) 2 = − 3 ∴ \therefore ∴ f ′ ( 0 ) = − 3 f' ( 0 ) = - 3 f ′ ( 0 ) = − 3 lim x → 1 + 0 f ′ ( x ) = 2 ⋅ 1 + 3 ( 1 − 1 ) 2 = 2 \displaystyle \lim\limits _{x \rightarrow 1 + 0} {} f' ( x ) = 2 \cdot 1 + 3 ( 1 - 1 ) ^{2} = 2 x → 1 + 0 lim f ′ ( x ) = 2 ⋅ 1 + 3 ( 1 − 1 ) 2 = 2 lim x → 1 − 0 f ′ ( x ) = 2 ⋅ 1 − 3 ( 1 − 1 ) 2 = 2 \displaystyle \lim\limits _{x \rightarrow 1 - 0} {} f' ( x ) = 2 \cdot 1 - 3 ( 1 - 1 ) ^{2} = 2 x → 1 − 0 lim f ′ ( x ) = 2 ⋅ 1 − 3 ( 1 − 1 ) 2 = 2 ∴ \therefore ∴ f ′ ( 1 ) = 2 f' ( 1 ) = 2 f ′ ( 1 ) = 2 f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1 f' ( 0 ) + f' ( 1 ) = - 3 + 2 = - 1 f ′ ( 0 ) + f ′ ( 1 ) = − 3 + 2 = − 1