f(x)=ax^3 +bx^2 +c f ′(x)=3ax^2 +2bx f(0)=−2 ⇒a(0)^3 +b(0)^2 +c=−2 ⇒c=−2 f(−2)=2 f ′(−2)=0 ⇒ { ((a(−2)^3 +b(−2)^2 −2=2)),((3a(−2)^2 +2b(−2)=0)) :} ⇒ { ((−8a+4b=4)),((12a−4b=0)) :} −8a+12a+4b−4b=4 ⇒4a=4 ⇒a=1 −8a+4b=4 ⇒b=1+2a ⇒b=1+2 ⇒b=3 f(x)=x^3 +3x^2 −2 f ′(x)=3x^2 +6x f(0)=(0)^3 +3(0)^2 −2=−2 vrai f(−2)=(−2)^3 +3(−2)^2 −2=−8+12−2=2 vrai f ′(−2)=3(−2)^2 +6(−2)=12−12=0 vrai


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Question Number 130331 by ayoubbacmath0 last updated on 24/Jan/21
$f(x)=ax^3 +bx^2 +c f ′(x)=3ax^2 +2bx f(0)=−2 ⇒a(0)^3 +b(0)^2 +c=−2 ⇒c=−2 f(−2)=2 f ′(−2)=0 ⇒ { ((a(−2)^3 +b(−2)^2 −2=2)),((3a(−2)^2 +2b(−2)=0)) :} ⇒ { ((−8a+4b=4)),((12a−4b=0)) :} −8a+12a+4b−4b=4 ⇒4a=4 ⇒a=1 −8a+4b=4 ⇒b=1+2a ⇒b=1+2 ⇒b=3 f(x)=x^3 +3x^2 −2 f ′(x)=3x^2 +6x f(0)=(0)^3 +3(0)^2 −2=−2 vrai f(−2)=(−2)^3 +3(−2)^2 −2=−8+12−2=2 vrai f ′(−2)=3(−2)^2 +6(−2)=12−12=0 vrai$
$f (x) = a x^{3} + {bx}^{2} + c$ $f^{'} (x) = 3 a x^{2} + 2 bx$ $f (0) = - 2$ $\Rightarrow a {(0)}^{3} + b {(0)}^{2} + c = - 2$ $\Rightarrow c = - 2$ $f (- 2) = 2$ $f^{'} (- 2) = 0$ $\Rightarrow {\begin{cases} a {(- 2)}^{3} + b {(- 2)}^{2} - 2 = 2 \\ 3 a {(- 2)}^{2} + 2 b (- 2) = 0 \end{cases}$ $\Rightarrow {\begin{cases} - 8 a + 4 b = 4 \\ 12 a - 4 b = 0 \end{cases}$ $- 8 a + 12 a + 4 b - 4 b = 4$ $\Rightarrow 4 a = 4$ $\Rightarrow a = 1$ $- 8 a + 4 b = 4$ $\Rightarrow b = 1 + 2 a$ $\Rightarrow b = 1 + 2$ $\Rightarrow b = 3$ $f (x) = x^{3} + {3 x}^{2} - 2$ $f^{'} (x) = {3 x}^{2} + 6 x$ $f (0) = {(0)}^{3} + 3 {(0)}^{2} - 2 = - 2 vrai$ $f (- 2) = {(- 2)}^{3} + 3 {(- 2)}^{2} - 2 = - 8 + 12 - 2 = 2 vrai$ $f^{'} (- 2) = 3 {(- 2)}^{2} + 6 (- 2) = 12 - 12 = 0 vrai$
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