Calc Proof? help?

Question

Consider the cubic function f(x)=ax^3+bx^2+cx+d where a doesnt not equal zero. Show that f can have zero, one, or two critical numbers and give an example of each case.
well im pretty sure you have to take the derivative first, so it would be 3ax^2+ 2bx+c
and then...do i do the quadratic formula to find zeros? idk! help!!

yea but how did u get those answers?
did u use the quadr. formula?
how>? im confused.

Answer 1

Yes you do need the quadratic formula.
The formula is [-B +/- sqrt(B^2 - 4AC)]/2A. I used capital letters here to distinguish these variables from the constants in the problem.
There are three general cases of solutions to the quadratic formula, which works out well since you are looking for three different solutions.
Case I: B^2 < 4AC. When the term inside the square root is negative, you cannot evaluate the whole equation, so there are no solutions (i.e. no critical numbers).

Case II: B^2 = 4AC. When this is true, the square root term evaluates to zero, and you are left with the critical number being -B/2A. In your quadratic function, A = 3a, B = 2b, and C = c. Making these substitutions gives
(2b)^2 = 4 * 3a * c
b^2 = 3ac.
Pick any set of three numbers that fits this criteria by picking two of the values, and figuring out what the third needs to be.
Example: when a=1 and b=1, c has to equal 1/3.
a=1 b=1 c=1/3

Case III: B^2 > 4AC. In this case your solutions are [-B +/- SQRT(B^2 - 4AC)]/AC. Let's do the same substitutions as in Case III:
(2b)^2 = 4 * 3a *c
b^2 > 3ac.
This time let's try a=1 and c=1. b^2 has to be greater than 3. b=2 works
a=1 b=2 c=1

Answer 2

By critical numbers I suppose you mean points where f has a minimum or a maximum. f is a polynomial function, so it's differentiable everywhere on the reals. We know it's derivative is f'(x) = 3ax^2 + 2bx + c. From Calculus, we know that if u is a local maximum or a local minimu of f, than f'(u) = 0, This is a necessary condition.

f'' is a polynomial of the second degree, so it can have 0 ,1 or 2 real roots. Then, we have:

1) f' has no real roots. In this case, f' is either strictly positive or strictly negative. This implies f is either strictly increasing or strictly decreasing and has no critical points.

2) f' has just one real root. In this case, f' is a constant multiplied by a perfect square, that is f'(x) = a(x-r)^2, where r is it's only root. We have f''(x) = 2a(x-r) and f''(r) = 0. And f'''(x) = 2a, which is constant and has the same sign of a. Since 3 (3rd derivative) is odd, r is an inflection point, but it's neither a maximum nor a minimum. f has no critical points.

3) f' has 2 real roots and f'' doesnt vanish at any of such roots. .This implies one of these roots is a maximum and the other is a minimum. So, f has 2 critical points.

We see f can have none or 2 critical points, but never jut one critical point. This is easy to see, because since f is a 3-degree-polynomial and 3 is odd, when x goes to -oo f(x) goes to - oo or + oo and x goes to oo f(x) goes to + oo or - oo. So, its impossible that f has just one crtical point.

Well, if you consider inflection points as critical points, then f can have 3 critical points.

Answer 3

for finding the critical numbers you need to put the derivative equal to 0 that is 3ax^2+ 2bx+c=0

This is a quadratic equation. It has Discriminant = (2b)^2 - 4*(3a)*c. If the discriminant is equal to zero the the critical numbers equal 1 as both the roots are same.

If the discriminant is less than 0 then the roots are imaginary and the number of critical numbers is zero

If the discriminant is more than 0 then the roots are real and different and the number of critical numbers equal 2

Answer 4

f'(x)=3ax^2+2bx+c
so
you need to find examples where
f'(x) =0 has no solutions, one solutions or 2 solutions

no solutions:
a=1, b=0 c=1,
f'(x)=3x^2+1>=1 for all x
1 solution:
a=1/3, b=-1 c=1
f'(x)=x^2-2x+1= (x-1)^2 = 0, so there is only one solution:x=1
2 solutions:
a=1/3, b=-3/2, c=2
f'(x)=x^2-3x+1=(x-1)(x-2)=0, so there are 2 solutions, x=1 and x=2

you can use quadratic formula, or simply what you know about second degree polynomials,
for example,
what i did was to take (x-1)(x-2) multiply them out, then i get
x^2-3x+2, and just equal the coefficients with
the coefficients of f'(x)= 3ax^2 +2bx +c,
so 3a=1, -3=2b and 2=c

remember you are just finding examples, not all the solutions to a particular problem, so you can play a bit. If you use the quadratic formula, you are going to get the discriminant:
4b^2-4(3a)c = 4(b^2-3ac), so
no solutions means finding a,b,c such that
b^2-3ac <0 (and for this you just play with numbers...., like b=0 and a=1=c)
etc

Answer 5

critical nuimbers arer solultions of the equation f'=0

your f' is 3ax^2+ 2bx+c = 0

this is a quadratic equation the solutions of this are given by the so called abc formula.

From this parabole it is known that it can hove none one or two zeros.

and that is your proof.

Extra info i give you is this : ( but thisis not needed )
x1,2 = { -2b + or - sqrt((2b)^2 - 4*3a*c) } / 2*3a

that is all

Answer 6

derive a quadratic equation from a cubic expresion

Answer 7

prepare right here: ??(a² - u²) du = ½u ?(a² - u²) + ½a² arcsin(u/a) + C this could nicely be a Trigonometric Substitution project: ?(a² - b²x²) ? x = (a/b)sin? ??(a² - u²) du = ??(a² - (asin?)²)acos? d? = ??(a² - a²sin²?)acos? d? = ??(a²(a million - sin²?))acos? d? = ??(a²cos²?)acos? d? = ?(acos?)acos? d? = ?(a²cos²?) d? = a²?(a million - sin²?) d? = a²[? - ½(? - cos?sin?)] + C = a²[arcsin(u/a) - ½(arcsin(u/a) - (?(a² - u²)/a)u/a)] + C = a²[arcsin(u/a) - ½arcsin(u/a) + ½u?(a² - u²)/a²] + C = a²[½arcsin(u/a) + ½u?(a² - u²)/a²] + C = ½a² arcsin(u/a) + ½u?(a² - u²) + C [This has been shown] u = asin? ? sin? = u/a = O/H ? ? = arcsin(u/a) du = acos? d? cos? = A/H = ?(a² - u²)/a