help on integration question!

[y3lloj3llo]

the question is:

two graphs have gradient functions dy/dx = 3x^2 + 3x + a and dy/dx = 3x^2 - 2x +1. the graphs cross at the point (1,a) and also where x= -2. find equations of the two graphs, and the value of a

i tried to solve by integrating both functions and substituting x and y in the first function with 1 and a respectively. that got me a y-intercept of -5/2. then i found c in the second integrated function in terms of a because i wasn't sure how to find it and then i equated the two together by subbing all x values with -2 like this:

(-2)³ + 3/2(-2)² - 2a - 5/2 = (-2)³ - (-2)² + (-2) + (a-1)

and i got that -3a = -21/3 so a = 7/2 and then to find the c of the second function, i added 1 to get c and i got 9/2 HOWEVER when i put the new functions into wolframalpha i got a graph that does not intersect at (1, 7/2). please help!!!!!!!!!!!!!!!!!

 

[yoscar04]

You got something wrong. I did something similar to what you did but got a=11/6. The corresponding plot is

Please include your detailed computation to see what went wrong with it.

 

[JeffM]

You can do this problem analytically though yoscar's solution is more elegant.

When you do your integrations (good first step), you get two constants of integration.

[MATH]f(x) = x^3 + 1.5x^2 + ax + c \text { and } \\ g(x) = x^3 - x^2 + x + d[/MATH] by integration.

Now we have three parameters to solve for. To do so, we need three equations. Basic stuff.

OK, you went to f(1) = a. Good idea. That should have got you

[MATH]f(1) = a \implies 1^3 + 1.5 * 1^2 + a * 1 + c = a \implies c = - 2.5.[/MATH]
And that does imply that the y-intercept of f(x) is - 5/2 = - 2.5. But so what? You have no reason to believe that g(x) has the same y-intercept.

I think you lost your way because you ignored the clue that g(1) = a also.

Solving that gets you that d = a - 1.

So we can restate the functions as

[MATH]f(x) = x^3 + 1.5x^2 + ax - 2.5 \text { and}\\ g(x) = x^3 - x^2 + x + a - 1.[/MATH]Now there are a number of ways to go based on the fact that f(-2) = g(-2).

Give it a go.

 

[y3lloj3llo]

You can do this problem analytically though yoscar's solution is more elegant.

When you do your integrations (good first step), you get two constants of integration.

[MATH]f(x) = x^3 + 1.5x^2 + ax + c \text { and } \\ g(x) = x^3 - x^2 + x + d[/MATH] by integration.

Now we have three parameters to solve for. To do so, we need three equations. Basic stuff.

OK, you went to f(1) = a. Good idea. That should have got you

[MATH]f(1) = a \implies 1^3 + 1.5 * 1^2 + a * 1 + c = a \implies c = - 2.5.[/MATH]
And that does imply that the y-intercept of f(x) is - 5/2 = - 2.5. But so what? You have no reason to believe that g(x) has the same y-intercept.

I think you lost your way because you ignored the clue that g(1) = a also.

Solving that gets you that d = a - 1.

So we can restate the functions as

[MATH]f(x) = x^3 + 1.5x^2 + ax - 2.5 \text { and}\\ g(x) = x^3 - x^2 + x + a - 1.[/MATH]Now there are a number of ways to go based on the fact that f(-2) = g(-2).

Give it a go.

yeah i did pretty much what you did where i found the intercept of the second function in terms of a and equated them both to one another when x= -2 but i keep getting a= 7/2!! this is what i did:

[MATH] (-2)^2 + 1.5(-2)^2 - 2a - 2.5 = (-2)^3 - (-2)^2 - 2 + a - 1 [/MATH][MATH]- 8 + 6 - 2.5 - 2a = -8 - 4 -2 + a - 1[/MATH][MATH]- 9/2 - 2a = - 15 + a [/MATH]
there must be something wrong on my part but i can't find what's wrong!! like i'm literally just doing the same thing over and over again and i can't identify where i'm going wrong!! sorry for any inconveniences and thanks for your help so far!!!

 

[JeffM]

yeah i did pretty much what you did where i found the intercept of the second function in terms of a and equated them both to one another when x= -2 but i keep getting a= 7/2!! this is what i did:

[MATH] (-2)^2 + 1.5(-2)^2 - 2a - 2.5 = (-2)^3 - (-2)^2 - 2 + a - 1 [/MATH][MATH]- 8 + 6 - 2.5 - 2a = -8 - 4 -2 + a - 1[/MATH][MATH]- 9/2 - 2a = - 15 + a [/MATH]
there must be something wrong on my part but i can't find what's wrong!! like i'm literally just doing the same thing over and over again and i can't identify where i'm going wrong!! sorry for any inconveniences and thanks for your help so far!!!

Assuming your algebra is correct, you are virtually done. Solve for a.

[MATH]\therefore 3a = 15 - \dfrac{9}{2} = \dfrac{21}{2} \implies a = 3.5 \implies[/MATH]
[MATH]f(x) = x^3 + 1.5x^2 + 3.5x - 2.5 \text { and } g(x) = x^3 - x^2 + x + 2.5.[/MATH]
Now check.

[MATH]f(1) = 1^3 + 1.5(1^2) + 3.5(1) - 2.5 = 3.5 = a. \ \checkmark.[/MATH]
[MATH]g(1) = 1^3 - 1^2 + 1 + 2.5 = 3.5 = a. \ \checkmark.[/MATH]
[MATH]f(2) = (-2)^3 + 1.5(-2)^2 + 3.5(-2) - 2.5 =\\ -8 + 4 * 1.5 - 7 - 2.5 = 6 - 17.5 = - 11.5 \text { and } \\ g(2) = (-2)^3 - (-2)^2 + (-2) + 2.5 = \\ - 8 - 4 - 2 + 2.5 = 2.5 - 14 = -11.5.\\ \therefore f(-2) = g(-2). \ \checkmark.[/MATH]
You needed to find three equations to solve for three unknowns. First year algebra. You got lost in the weeds.

 

[y3lloj3llo]

Assuming your algebra is correct, you are virtually done. Solve for a.

[MATH]\therefore 3a = 15 - \dfrac{9}{2} = \dfrac{21}{2} \implies a = 3.5 \implies[/MATH]
[MATH]f(x) = x^3 + 1.5x^2 + 3.5x - 2.5 \text { and } g(x) = x^3 - x^2 + x + 2.5.[/MATH]
Now check.

[MATH]f(1) = 1^3 + 1.5(1^2) + 3.5(1) - 2.5 = 3.5 = a. \ \checkmark.[/MATH]
[MATH]g(1) = 1^3 - 1^2 + 1 + 2.5 = 3.5 = a. \ \checkmark.[/MATH]
[MATH]f(2) = (-2)^3 + 1.5(-2)^2 + 3.5(-2) - 2.5 =\\ -8 + 4 * 1.5 - 7 - 2.5 = 6 - 17.5 = - 11.5 \text { and } \\ g(2) = (-2)^3 - (-2)^2 + (-2) + 2.5 = \\ - 8 - 4 - 2 + 2.5 = 2.5 - 14 = -11.5.\\ \therefore f(-2) = g(-2). \ \checkmark.[/MATH]
You needed to find three equations to solve for three unknowns. First year algebra. You got lost in the weeds.

oh my gosh thank you!!! i got this at the beginning and it turns out i put the wrong values into wolframalpha LMAO

thank you so much for your help!!!! sorry for any inconveniences!!!