The function f(X) = AX^3 + BX^2 + CX is 0 at X = -2....

[kb692006]

The function f(X) = AX^3 + BX^2 + CX is 0 at X = -2 [A(-2;0)] and has a tangent y = (1/2)X + 1 in point B(2;2). How big are A, B, C, (which are elements of R)?

I have gotten that B is equal to 1/4.

I know that I need to flip the first derivative (?) of f(X) --> f'(X) = 1/2 since the tagent is y = mx + b and m = 1/2, which makes f'(2) = 1/2, but I seem to have a flaw. When I break it down, I get:

f(X) = AX^3 + BX^2 + CX

POINT A(-2;0)

I. f(-2) = -8A + 4B - 2C = 0

POINT B(2;2)

II. f(2) = 8A + 4B + 2C = 2

I + II => 8B = 2 => B = 1 / 4

tangent y= 1/2*X + 1 which touches it in point B(2;2)

The tangent of a function always has the same gradient as the function. The basic form of a tangent is y = mx + b, where "m" is the gradient HERE y=1/2*X+1

The gradient of a function is its derivative ---> applying the point B ---> f'(2) = 1/2

f'(x) = 3AX^2 + 2BX + C

applying point B, m, and my result for B.

f'(2) = 12A + 4B + C = 1/2

III. f'(2) = 12A + 1 + C = 1/2

III. f'(2) = 12A + 1 + C = 1/2 ///*2
+
I. f(-2) = -8A + 1 - 2C = 0
=
16A + 2 = 1 <=> 16A = -1 <=> A = -1 / 16

But this result for A is WRONG. The solution should be A = -1/8

I'm hoping somebody can help me find my error. Thank you!

 

[stapel]

Your statement of the exercise makes it look like A, B, and C are points, which doesn't make much sense to me. Please clarify: What do you mean when you say that the coefficient A is "A(-2:0)"? And so forth.

Thank you.

Eliz.

 

[kb692006]

Your statement of the exercise makes it look like A, B, and C are points, which doesn't make much sense to me. Please clarify: What do you mean when you say that the coefficient A is "A(-2:0)"? And so forth.

Thank you.

Eliz.

it says the function f(X) = A*X³ + B*X² + C*X is 0 at x=-2 ( as in, f(-2)=0 ). ---> POINT A is (-2;0) [f(x);y]

f(X) also has a tangent y= 1/2*X + 1 which touches it in point B(2;2).

i wanna know the way to A and C

 

[stapel]

So the coefficients of the function f are indeed points...? So the function is "f(x) = (-2, 0)x³ + (2, 2)x² + Cx, and you're needing to find the x- and y-coordinates of the point C...?

Thank you.

Eliz.

 

[kb692006]

So the coefficients of the function f are indeed points...? So the function is "f(x) = (-2, 0)x³ + (2, 2)x² + Cx, and you're needing to find the x- and y-coordinates of the point C...?

Yes please

 

[soroban]

Hello, kb692006!

Your reasoning and your work is correct . . . until the end.

The function $\displaystyle f(x) \:= \:Ax^3\,+\,BX^2\,+\,CX$ is $\displaystyle 0$ at $\displaystyle x\,=\,-2:\;A(-2,0)$
and has a tangent $\displaystyle y \:= \:\frac{1}{2}x\,+\,1$ at point $\displaystyle B(2,2).$
Find $\displaystyle A,]\,B,\,C$

I have gotten that $\displaystyle B \,=\,\frac{1}{4}$

I know that I need to flip the first derivative (?) of f(X) --> f'(X) = 1/2
since the tagent is y = mx + b and m = 1/2, which makes f'(2) = 1/2,
but I seem to have a flaw. When I break it down, I get:

. . $\displaystyle f(x)\:=\:Ax^3\,+\,Bx^2\,+\,Cx$

Point $\displaystyle A(-2,0):\;\;I:\:f(-2)\:=\:-8A\,+\,4B\,-\,2C\:=\: 0$

Point $\displaystyle B(2,2):\;\;II:\:f(2)\:=\:8A\,+\,4B\,+\,2C\:=\:2$

$\displaystyle I\,+\,II\;\Rightarrow\;8B\,=\,2\;\Rightarrow\;B\,= \frac{1}{4}$


Tangent: $\displaystyle y\:=\:\frac{1}{2}x\,+\,1$ which touches $\displaystyle f(x)$ at point $\displaystyle B(2,2).$

The tangent of a function always has the same gradient as the function.
The basic form of a tangent is: $\displaystyle y \:=\: mx\,+\,b$, where $\displaystyle m$ is the gradient HERE: $\displaystyle \,y\:=\:\frac{1}{2}x\,+\,1$

The gradient of a function is its derivative.
. . Applying the point B $\displaystyle \rightarrow\;\;f'(2)\,=\,\frac{1}{2}$

. . \(\displaystyle f'(x) \:= \:3Ax^2\,+\.2Bx\,+\,C\)

Applying point B, m, and my result for B: $\displaystyle \,f'(2) :=\:12A\,+\,4B\,+\,C\:=\:\frac{1}{2}$

. . $\displaystyle III.\;f'(2)\:=\:12A\,+\,1\,+\,C\:=\:\frac{1}{2}$


We have: $\displaystyle \:\begin{array}{cc}f'(2) & = & 12A\,+\,1\,+\,C & = & \frac{1}{2}\\f(-2)& = &-8A\,+\,1\,-\,2C & = & 0\end{array}$

. . $\displaystyle 16A\,+\,2\:=\:1\;$ < --- Here!

The first equation is: $\displaystyle \:12A\,+\,C\:=\:-\frac{1}{2}$ [1]

The second is: $\displaystyle \:-8A\,-\,2C\:=\:-1\;\;\Rightarrow\;\;-4A\,-\,C\:=\:-\frac{1}{2}$ [2]

Add [1] and [2]: \(\displaystyle \:8A\:=\:-1\;\;\Rightarrow\;\;\fbox{A \:=\:-\frac{1}{8}}\)

 

[kb692006]

Hello, kb692006!

Your reasoning and your work is correct . . . until the end....

Thank You i appreciate it so that is the way to A &C

it says the function f(X) = A*X³ + B*X² + C*X is 0 at x=-2 ( as in, f(-2)=0 ). ---> POINT A is (-2;0) [f(x);y]

f(X) also has a tangent y= 1/2*X + 1 which touches it in point B(2;2).

i wanna know the way to A and C.