How Do You Solve for Constants in a Cubic Function with Horizontal Tangents?

[mathaTon]

If I am asked to find the constants a, b, c, d such that the graph of
f(x)= ax^3 + bx^2 + cx+ d has horizontal tangent lines at the points (-2, 1) and (0, -3).

I am not sure what to go about doing it...is that asking me to find what? (I don't know what do the constant stand for? in form of y= mx+b?)

I know though, the first thing I would do is, find the derivative of the function..."f(x)= ax^3 + bx^2 + cx+ d "

which would be...

f' (x) = 3 ax^2+ 2 bx+ c
then sub the value of x? x = (-2) into the last equation..
which will equal to
12 a- 4b+ c

nowwwwww? what do I do next?


and I have another q.


Given h= f 0 g, g(3)=7, g'(3)=4, f(2)=4, f'(7)=-6.
now how do I determine the h' (3)?

again i am half way through the answer...

I think thinking of solving it with product rule??
h(x)= f(g) x)) h(x)= f' (g(x) g'(x)

h' (x)= f' (g (3)= g'3= f' (7) (4) = (-6) (-4) = -24


soooo please help?

 

[chroot]

You've done the first problem correctly so far. You know the tangent line is horizontal when the derivative is zero, so you know the equation you found:

f' (x) = 3 ax^2+ 2 bx+ c

must be zero at x=-2. You also know the original equation,

f(x)= ax^3 + bx^2 + cx+ d

must be 1 at x=-2, as given.

You also have another set of equations, with f'(x) = 0 and f(x) = -3 at x=0.

Solve the simultaneous equations.

- Warren

 

[mathaTon]

You've done the first problem correctly so far. You know the tangent line is horizontal when the derivative is zero, so you know the equation you found:

f' (x) = 3 ax^2+ 2 bx+ c

must be zero at x=-2. You also know the original equation,

f(x)= ax^3 + bx^2 + cx+ d

must be 1 at x=-2, as given.

You also have another set of equations, with f'(x) = 0 and f(x) = -3 at x=0.

Solve the simultaneous equations.

- Warren

Hi Warren, thanks for replying..
But I don't get the part where you think _ must be zero ( do u mean the x value ?)

and how do u figure the value of 1 @ x= -2?

i am lost..I m taking a summer course it is going at fast as speed of the light..it is hard to absorb the material.

 

[Office_Shredder]

If there is a horizontal tangent at a point P=(x,y), then f '(x)=0 (the derivative)

 

[Office_Shredder]

Given h= f 0 g, g(3)=7, g'(3)=4, f(2)=4, f'(7)=-6.
now how do I determine the h' (3)?

again i am half way through the answer...

I think thinking of solving it with product rule??
h(x)= f(g) x)) h(x)= f' (g(x) g'(x)

h' (x)= f' (g (3)= g'3= f' (7) (4) = (-6) (-4) = -24

Take it a bit slower... you seem to be missing some vital parentheses h'(x) = f '(g(x))*g'(x). So for x=3:

f '(g(3))*g'(3)

Now try solving that

 

[mathaTon]

Take it a bit slower... you seem to be missing some vital parentheses h'(x) = f '(g(x))*g'(x). So for x=3:

f '(g(3))*g'(3)

Now try solving that

I feel like a loser.. oh well, let me give my best shot:shy:

Since h (x)= f (g (x))
then can I apply this ruling to solve for h (x)

f '(g(3))*g'(3) = 9? (by multiplying?)

 

[mathaTon]

Am I right?? anyone? Please

 

[Office_Shredder]

I feel like a loser.. oh well, let me give my best shot:shy:

Since h (x)= f (g (x))
then can I apply this ruling to solve for h (x)

f '(g(3))*g'(3) = 9? (by multiplying?)

f '(g(3)) = f '(7) from the info. f '(7) = -6. Multiply that by g'(3), and you get -24

You had it right the first time, I just couldn't tell because of the missing parentheses if you used the right method

 

[VietDao29]

Uhmm, ok, for the first problem:
You have 4 unknowns, namely a, b, c, and d. So you must need 4 equation to solve for them.
Horizontal tangent line has a slope of 0, right? Since it's parallel to the Ox axis.
So, at x = 0, and x = -2, the graph of f(x) has horizontal tangent lines. Or in other words:
f'(0) = f'(-2) = 0. Right?
This is the first 2 equations, right?
Given that the 2 points (-2, 1), and (0, -3) are on the curve. So you'll have another 2 equations:
f(-2) = 1, and f(0) = -3.
Having 4 equations, can you solve for the 4 unknowns?
Can you get this? Is there anything unclear? :)