Stationary and Inflection Points

Years

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 2

Find the variables $a, b, c$ $a,b,c$ in

f (x) = x^{3} +

$f(x)=x^3+$ $a$ $a$

x^{2} +

$x^2+$ $b$ $b$

x +

$x+$ $c$ $c$ ,
given that:

Stationary Point

x = 3

$x=3$

Inflection Point

(1, 4)

$(1,4)$

$a =$ $a=$ (-3)

$b =$ $b=$ (-9)

$c =$ $c=$ (15)

Question 2 of 2

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)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

A point that satisfies

$f'(x)=0$ is called a Stationary Point

A point that satisfies

$f”(x)=0$ is called an Inflection Point

First, 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 function

Inflection 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 $2$
$b$	$=$	$-3$

$a=1/2$

$b=-3$

$c=0$

$f(x)$	$=$	$x^3+ax^2+bx+c$
$f'(x)$	$=$	$3x^2+2ax+b=0$	Equate $f'(x)$ to $0$
$f'(3)$	$=$	$3(3)^2+2(-3)(3)+b=0$	Substitute $x=3$ and $a=-3$

Stationary and Inflection Points

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

2. Question

Hint

Quizzes

$6+2a$	$=$	$0$
$6+2a$ $-6$	$=$	$0$ $-6$	Subtract $6$ from both sides
$2a$	$=$	$-6$
$2a$ $-:2$	$=$	$-6$ $-:2$	Divide both sides by $2$
$a$	$=$	$-3$

$3(9)+2(-9)+b$	$=$	$0$
$27-18+b$	$=$	$0$
$9+b$	$=$	$0$
$9+b$ $-9$	$=$	$0$ $-9$	Subtract $9$ from both sides
$b$	$=$	$-9$

$f(x)$	$=$	$x^3+ax^2+bx+c$
$4$	$=$	$(1)^3+(-3)(1^2)+(-9)(1)+c$	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
$15$	$=$	$c$
$c$	$=$	$15$