- Problem
- Detailed Solution
- Summary Solution

### Determining parameters of a polynomial function

Let f be a third degree polynomial, \(f(x)=ax^3+bx^2+cx+d\).

Determine parameter values a, b, c, and d, so that f has a root and an inflection point at \({x_0}=0\), a local minimum or maximum at \({x_1}=-2\), and the tangent at \({x_2}=4\) has a slope of 3 .

### Preliminary remarks

To solve this problem, you need the following competences:

(i) to characterize a root of a function

(ii) to characterize an inflection point of a function

(iii) to characterize an extremum of a function

(iv) to characterize the slope of a tangent to the graph of a function

The theory behind all this is

- in block II (tangents), and
- in block III learning sequence 2: sections 6 (extrema) and 7 (inflection points)

of our script.

### Formal approach

We make use of the **first information**:

f has a root at \({x_0}=0\). This means that, if x is replaced by the number 0, f(x) is itself 0: \(f({x_0})=f(0)=0\).

Thus, we obtain

\(f(0)=a \cdot 0^3+b \cdot 0^2+c \cdot 0+d\).

Since a, b, and c are nullified by powers of 0, we have \(\;d=0\).

This is our first finding, and our function now reads \(\;f(x)=ax^3+bx^2+cx\), with a, b, and c still unknown.

We now make use of the **second information**:

f has an inflection point at \({x_0}=0\). Although we stay at the
same place on the x-axis, this information is quite different from
the first one. The implication is that the 2^{nd} order
derivative of f, that is f'', takes the value 0: \(f''(x)= 0\).

We determine the derivatives we need:

\(f(x)=ax^3+bx^2+cx+d \quad \Rightarrow \quad f'(x)=3ax^2+2bx+c \quad \Rightarrow \quad f''(x)=6ax+2b\)

We evaluate f'' for x=0:

\(f''(0)=6a \cdot 0+2b=0\), which means \(\;b=0\).

Now we have the second of our four paramater values, and we can state

\(f(x)=ax^3+cx\), with only *a* and *c* missing.

The **third information**

is the one on the extremum: since this is at \({x_1}=-2\), the necessary condition is that f' must be 0 there:

\(f'(-2)=0\)

Since \(\;f'(x)=3ax^2+2bx+c\;\) (remember that \(\; b=0\,\)), we have:

\(f'(-2)=3a \cdot (-2)^2+2b \cdot (-2)+c=12a-4b+c=0\)

We keep this equation that contains two unknown symbols, hoping that

the ** last information**

yields the "missing piece": since the slope of the tangent to the
graph of f is 3 for x=4, we make use of the fact that this is
nothing else but the value of the 1^{st} order derivative
of f at that place.

Therefore, \(\;f'(4)=3\,\), and with \(\;f'(x)=3ax^2+2bx+c \;\) we obtain

\( f'(4)=3a \cdot 4^2+2b \cdot 4+c=48a+c=3\).

Here again, we have one equation for two unknown symbols, but **taken
together**, these **two equations are enough**
to find the values of a and c:

We make use of \(12a+c=0\) (see above) to state that: \(c=-12a\).

Then, we replace c in our last equation by that multiple of a, and we obtain

\( 48a+(-12a)=3 \quad \Rightarrow \quad 36a=3 \quad \Rightarrow \quad a=\frac{1}{12}\)

And since c=−12a, we conclude that \(c=-1\).

This completes our results, and we write down the complete equation for f:

\(f(x)=\frac{1}{12}x^3-x\)

To end the exercise, we might visualize our findings by sketching
the graph, thus verifying whether it meets the four conditions set
up in the problem.

See the diagram to
do so.

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

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

\(f''(x)=6ax+2b\)

\(f'''(x)=6a\)

I: \(\quad f(0)=0 \quad \Rightarrow \quad d=0\)

II: \(\quad f''(0)=0 \quad \Rightarrow \quad b=0\)

III: \(\quad f'(-2)=0 \quad \Rightarrow \quad 12a-4b+c=0\)

IV: \(\quad f'(4)=3 \quad \Rightarrow \quad 48a+8b+c=3\)

Replacing \(b=0\;\) and \(\;d=0\;\) in equations III and IV, we have:

\(12a+c=0\) and \(48a+c=3 \quad \Rightarrow \quad a=\frac{1}{12}\) and \(c=-1\)

Solution: \(\quad f(x)=\frac{1}{12}x^3-x\)