Formula with 2 points of inflection

Question

$x^3$ has a point of inflection at $x=0$. How will you modify the formula to add a 2nd point of inflection at $x=1$?

Plot of $x^3$

Plot of $x^3(x-1)^3$

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Update

The plot I am aiming to achieve has a shape similar to the graph below. However I would like the 2 inflection points at $(0, 1)$ and $(1, 0.05)$, and intersects the x-axis at $(1.5,0)$ and y-axis at $(0,1)$.

Graph with similar target shape

Current attempt

The closest I can get is using $1 - [ 16x^3 - 23x^4 + 9x^5 ]$ using J.M.'s equation inside the $[]$ square brackets with $\alpha=2$ and $\beta=1$. How should I bring the point of inflection at $x=1$ up to around $y=0.05$? Its currently at $y=-1$

Answer 1

It seems that there is an implicit constraint (because else George V. Williams' answer (+1) would be perfect !) : that the derivatives at $x=0$ and $1$ are $0$ !

With these constraints I got a polynomial of degree $6$ $P(x)=\frac{706}{135}x^6-\frac{385}{18}x^5+\frac{5389}{180}x^4-\frac{3977}{270}x^3+1$
but the behavior was not nice near $\frac 32$ so that I tried again with a polynomial of degree $7$ and the additional constraint $P'\left(\frac 32\right)=-1$ (change this if wished !) : $$P(x)=a x^7+b x^6+c x^5+dx^4+ex^3+fx^2+gx+h$$ From $P(0)=1$ we get $\ h=1$
From $P(1)=\frac 1{20}$ we get $\ a+b+c+d+e+f+g=\frac 1{20}-1$
We want too $P\left(\frac 32\right)=0$ but will write this down later...
$$P'(x)=7ax^6+6b x^5+5c x^4+4dx^3+3ex^2+2fx+g$$ Should $P'(0)=P'(1)=0$ be wished then $g=0\ $ and $\ 7a+6b+5c+4d+3e+2f=0$
(we will impose too $P'\left(\frac 32\right)=-1$ at the end) $$P''(x)=42ax^5+30b x^4+20c x^3+12dx^2+6ex+2f$$

From $P''(0)=P''(1)=0$ we get $f=0\ $ and $\ 7a+5b+\frac {10}3c+2d+e=0$

Let's put everything together (the two last lines from the array concern $64P'\left(\frac 32\right)$ and $128P\left(\frac 32\right)$) : $$P(x)=a x^7+bx^6+c x^5+dx^4+ex^3+1$$ with \begin{array}{cccccc} a &+b &+c &+d &+e&= &-\frac{19}{20}\\ 7a &+6b &+5c &+4d &+3e &= &0\\ 7a &+5b &+\frac {10}3c&+2d&+e&=&0\\ 5103a &+ 2916b &+ 1620c &+ 864d &+ 432e &=&-64\\ 2187a &+ 1458 b &+ 972 c &+ 648d &+ 432e &=&-128\\ \end{array}

The solution deduced from this system is : $$P(x)=-\frac{838}{135}x^7+\frac{4477}{135}x^6-\frac{1223}{18}x^5+\frac{34603}{540}x^4-\frac{6491}{270}x^3+1$$

with the picture :

Answer 2

One rather general family is given by

$$2(5\alpha-2\beta)x^3+(7\beta-15\alpha)x^4+3(2\alpha-\beta)x^5$$

I obtained this through Hermite interpolation. That involves derivatives, so I'm not sure if this counts for a "precalculus" answer.

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I would like the 2 inflection points at $(0,1)$ and $(1,0.05)$, and intersects the x-axis at $(1.5,0)$ and y-axis at $(0,1)$.

Using Hermite interpolation again, we obtain the polynomial

$$1-\frac{3977}{270}x^3+\frac{5389}{180}x^4-\frac{385}{18}x^5+\frac{706}{135}x^6$$

You can easily verify that this has the properties needed.

Another possibility, whose shape is a bit nearer to what the OP seems to want, is

$$1-\frac{6491}{270}x^3+\frac{34603}{540}x^4-\frac{1223}{18}x^5+\frac{4477}{135}x^6-\frac{838}{135}x^7$$

Answer 3

A fourth degree polynomial can have two points of inflection. Let's pick an arbitrary one: $f(x) = ax^4 + bx^3 + cx^2 + dx + e$.

Inflection points occur when the second derivative is $0$: $f^{\prime\prime}(x) = 12ax^2 + 6bx^2 + c = 0$. We want these roots to occur at 0 and 1.

Substituting $x=0$, we get $c = 0$. Substituting $x=1$, we can pick $a = 1, b = -2$. Put these back into the original equation: $x^4 - 2x^3 + dx + e$. One solution is simply $x^4 - 2x^3$, but a prettier one is $x^4 - 2x^3 + x + \frac{1}{4}$

Answer 4

Use a fourth-degree polynomial $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, and get that:

$$ f''(x) = 12ax^2 + 6bx + c = 0 \text{ at } x = 0 \text{ and } 1$$

So we get that:

$$ c = 0 \text{ and } 12a + 6b = 0$$

Well that's still not enough! Based on what you said, we also know that $e = 1$ (since $f(0) = 1$). We also need that:

$$f(1) = a+b+c+d+e = \\ a + b + d + 1 = 0.05$$

So close, and your last criterion, that $f(1.5) = 0$, will let us finish it:

$$ 5.0625 a + 3.375 b + 1.5d + 1 = 0$$

We have the following system of equations:

$$ 2a + b = 0 \\ a + b + d + 1 = 0.05 \\ 5.0625 a + 3.375 b + 1.5d + 1 = 0 $$

Solving the system of equations, we get:

$$a = -\frac{34}{15} \\ b = \frac{68}{15} \\ d = -\frac{193}{60} $$

Graphing the result: