Years
>
Year 11>
Derivative Curve Sketching>
Stationary and Inflection Points>
Stationary and Inflection PointsStationary and Inflection Points
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 2 questions completed
Questions:
- 1
- 2
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- Answered
- Review
-
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` -
-
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` -