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AP Calculus BC>
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Stationary and Inflection Points>
Stationary and Inflection PointsStationary and Inflection Points
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Question 1 of 2
1. Question
Find the variables `a,b,c` in `f(x)=x^3+``a``x^2+``b``x+``c`,
given that:Stationary Point `x=3`Inflection Point `(1,4)`-
`a=` (-3)`b=` (-9)`c=` (15)
Hint
Help VideoCorrect
Excellent
Incorrect
A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f”(x)` to `0` and substitute `x=1` (from inflection point `(1,4)`)`f(x)` `=` `x^3+ax^2+bx+c` `f'(x)` `=` `3x^2+2ax+b` `f”(x)` `=` `6x+2a=0` Equate `f”(x)` to `0` `f”(1)` `=` `6(1)+2a=0` Substitute `x=1` `6+2a` `=` `0` `6+2a``-6` `=` `0``-6` Subtract `6` from both sides `2a` `=` `-6` `2a``-:2` `=` `-6``-:2` Divide both sides by `2` `a` `=` `-3` Next, equate `f'(x)` to `0` and substitute the stationary point `x=3` and `a=-3``f(x)` `=` `x^3+ax^2+bx+c` `f'(x)` `=` `3x^2+2ax+b=0` Equate `f'(x)` to `0` `f'(3)` `=` `3(3)^2+2(-3)(3)+b=0` Substitute `x=3` and `a=-3` `3(9)+2(-9)+b` `=` `0` `27-18+b` `=` `0` `9+b` `=` `0` `9+b``-9` `=` `0``-9` Subtract `9` from both sides `b` `=` `-9` Finally, substitute known values into the function to solve for `c``a=-3``b=-9`Inflection Point `(x,f(x))=(1,4)``f(x)` `=` `x^3+ax^2+bx+c` `4` `=` `(1)^3+(-3)(1^2)+(-9)(1)+c` Substitute known values `4` `=` `1+(-3)(1)-9+c` `4` `=` `1-3-9+c` `4` `=` `-11+c` `4``+11` `=` `-11+c``+11` Add `11` to both sides `15` `=` `c` `c` `=` `15` `a=-3``b=-9``c=15` -
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Question 2 of 2
2. Question
Find the variables `a,b,c` in `f(x)=``a``x^3+``b``x^2+``c``x-5`,
given that:Stationary Point `x=0`Inflection Point `(2,-13)`Write fractions in the format “a/b”-
`a=` (1/2)`b=` (-3)`c=` (0)
Hint
Help VideoCorrect
Well Done
Incorrect
A point that satisfies `f'(x)=0` is called a Stationary PointA point that satisfies `f”(x)=0` is called an Inflection PointFirst, equate `f'(x)` to `0` and substitute `x=0` (stationary point)`f(x)` `=` `ax^3+bx^2+cx-5` `f'(x)` `=` `3ax^2+2bx+c=0` Equate to `0` `f'(0)` `=` `3a(0)^2+2b(0)+c=0` Substitute `x=0` `0+0+c` `=` `0` `c` `=` `0` Next, equate `f”(x)` to `0` and substitute `x=2` (from inflection point `(2,-13)`)`f(x)` `=` `ax^3+bx^2+cx-5` `f'(x)` `=` `3ax^2+2bx+c` `f”(x)` `=` `6ax+2b=0` Equate to `0` `f”(2)` `=` `6a(2)+2b=0` Substitute `x=2` `12a+2b=0` Form another equation by substituting known values into the functionInflection Point `(x,f(x))=(2,-13)``c=0``f(x)` `=` `ax^3+bx^2+cx-5` `-13` `=` `a(2^3)+b(2^2)+(0)(2)-5` Substitute known values `-13` `=` `8a+4b+0-5` `-13``+13` `=` `8a+4b+0-5``+13` Add `13` to both sides `0` `=` `8a+4b+8` `0``-:2` `=` `8a+4b+8``-:2` Divide both sides by `2` `0` `=` `4a+2b+4` `4a+2b+4` `=` `0` Use elimination method on the two equations to solve for `a`\begin{matrix}
\: & \color{#9a00c7}{12a} & \color{#9a00c7}{+} & \color{#9a00c7}{2b} & \: & \: & \color{#9a00c7}{=} & \color{#9a00c7}{0} \\
– & \color{#9a00c7}{4a} & \color{#9a00c7}{+} & \color{#9a00c7}{2b} & \color{#9a00c7}{+} & \color{#9a00c7}{4} & \color{#9a00c7}{=} & \color{#9a00c7}{0} \\
\hline
\: & 8a & + & 0 & – & 4 & = & 0
\end{matrix}`8a-4` `=` `0` `8a` `=` `4` `8a``-:8` `=` `4``-:8` Divide both sides by `8` `a` `=` `1/2` Substitute the value of `a` to one of the equations and solve for `b``12a+2b` `=` `0` `12(1/2)+2b` `=` `0` Substitute `a=1/2` `6+2b` `=` `0` `2b` `=` `-6` `2b``-:2` `=` `-6``-:2` Divide both sides by `2` `b` `=` `-3` `a=1/2``b=-3``c=0` -