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if f(x) = -x^3 +3x^2 -2, find the coordinates of the point(s) where the tangent line is horizontal

Guest Nov 20, 2018

Best Answer 

 #1
avatar+20680 
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if f(x) = -x^3 +3x^2 -2, find the coordinates of the point(s) where the tangent line is horizontal

 

\(\begin{array}{|lrcllcl|} \hline & f(x) &=& -x^3+3x^2-2 \\\\ & \boxed{f'(x) = 0\ ?} \\\\ & \boxed{f'(x) = -3x^2+6x } \\ & -3x^2+6x &=& 0 \\ & x(-3x+6) &=& 0 \\ 1. & x &=& 0, \qquad & y &=& -0^3+3\cdot 0^2 - 2 \\ & && & y &=& -2 \\ & \text{Point}_1~ (0,-2) \\\\ 2. & -3x+6 &=& 0 \\ & 3x &=& 6 \\ & x &=& \dfrac{6}{3} \\ & x &=& 2, \qquad & y &=& -2^3+3\cdot 2^2 - 2 \\ & && & y &=& -8+12-2 \\ & && & y &=& 2 \\ & \text{Point}_2~ (2,2) \\ \hline \end{array}\)

 

 

laugh

heureka  Nov 20, 2018
 #1
avatar+20680 
+10
Best Answer

if f(x) = -x^3 +3x^2 -2, find the coordinates of the point(s) where the tangent line is horizontal

 

\(\begin{array}{|lrcllcl|} \hline & f(x) &=& -x^3+3x^2-2 \\\\ & \boxed{f'(x) = 0\ ?} \\\\ & \boxed{f'(x) = -3x^2+6x } \\ & -3x^2+6x &=& 0 \\ & x(-3x+6) &=& 0 \\ 1. & x &=& 0, \qquad & y &=& -0^3+3\cdot 0^2 - 2 \\ & && & y &=& -2 \\ & \text{Point}_1~ (0,-2) \\\\ 2. & -3x+6 &=& 0 \\ & 3x &=& 6 \\ & x &=& \dfrac{6}{3} \\ & x &=& 2, \qquad & y &=& -2^3+3\cdot 2^2 - 2 \\ & && & y &=& -8+12-2 \\ & && & y &=& 2 \\ & \text{Point}_2~ (2,2) \\ \hline \end{array}\)

 

 

laugh

heureka  Nov 20, 2018

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