calculus: let f(x)=ax³+bx²+cx+d. if b²=3ac show that f(x) has only one turning point.

[markosheehan]

calculus: let f(x)=ax³+bx²+cx+d. if b²=3ac show that f(x) has only one turning point.

[FONT=&quot]let f(x)=ax³+bx²+cx+d. if b²=3ac show that f(x) has only one turning point. how do you do this[/FONT]

 

[Deleted member 4993]

let f(x)=ax³+bx²+cx+d. if b²=3ac show that f(x) has only one turning point. how do you do this

What is the mathematical definition of a turning point?

You are given a cubic function.

How many turning points does a cubic function have, in general?

What does the graph of a cubic polynomial look like with only one turning point?

 

[Ishuda]

let f(x)=ax³+bx²+cx+d. if b²=3ac show that f(x) has only one turning point. how do you do this

Admittedly it has been a long time since I have taken a class in calculus but I think I remember something like the following: If
(1) \(\displaystyle f^{(j)}(x_0)\, =\, 0\), j = 1, 2, 3, ..., n-1 \(\displaystyle \ge\) 1.
(2) \(\displaystyle f^{(n)}(x_0)\, \ne\, 0\).
then
(i) If n is even then x₀ is a turning point with \(\displaystyle f(x_0)\) a minimum if \(\displaystyle f^{(n)}(x_0)\, \gt\, 0\) and a maximum if \(\displaystyle f^{(n)}(x_0)\, \lt\, 0\)
(ii) Otherwise x₀ is a point of inflection and not a turning point.

If this is the case, then given f(x)=ax³+bx²+cx+d; if b²=3ac, how many turning points does f(x) have? Seems to me it is zero.

 

[soroban]

Let \(\displaystyle f(x)=ax^3+bx^2+cx+d.\)
If \(\displaystyle b^2=3ac\), show that \(\displaystyle f(x)\) has only one turning point.

To find the turning points, solve \(\displaystyle f'(x) = 0.\)

We have: .\(\displaystyle 3ax^2 + 2bx + c \:=\:0\)

Quadratric Formula: .\(\displaystyle x \;=\;\dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a} \;=\;\dfrac{-b \pm \sqrt{b^2-3ac}}{3a} \)

If \(\displaystyle b^2 = 3ac\), the discriminant is zero and the quadratic has one root.

 

[Ishuda]

To find the turning points, solve \(\displaystyle f'(x) = 0.\)

We have: .\(\displaystyle 3ax^2 + 2bx + c \:=\:0\)

Quadratric Formula: .\(\displaystyle x \;=\;\dfrac{-2b \pm \sqrt{4b^2 - 12ac}}{6a} \;=\;\dfrac{-b \pm \sqrt{b^2-3ac}}{3a} \)

If \(\displaystyle b^2 = 3ac\), the discriminant is zero and the quadratic has one root.

One root at x₀=-b/(3a). So, looking at
f''(x) = 6 a x + 2b
and evaluating at x₀ we have
f''(x₀) = 0
Thus look at
f''' = 6 a
and we find that the point x₀ is a point of inflection, not a turning point? So, possibly, as mentioned above, there are zero turning points [just highlight the last line in the above post to see 'Seems to me it is zero.']