Find a cubic function given the inflection point and minimal point

Question

I came across this task that I just couldn't figure out.

The task gave me an extremum(1,1) and the inflection point(2,3), and I need to figure out the cubic function given the values.

Assuming the cubic function would look like f(x)=ax^3+bx^2+cx+d, the following is what I've done so far:

a+b+c+d = 1 (y for extremum)
8a+4b+2c+d = 3 (y for inflection point)

Derivative function's 0 for the extremum:

3a+2b+c = 0

Second derivative's 0 for the inflection point:

12a+2b=0

After this, I just can't figure out, so any help would be appreciated.

Answer 1

Write all the variables in terms of $a$ and solve as follows:

$$12a + 2b = 0 \implies b = -6a$$

$$\therefore \, 3a + 2(-6a) + c = 0 \implies c = 9a$$

$$\therefore \, a + (-6a) + (9a) + d = 1 \implies d = 1 - 4a$$

$$\therefore \, 8a + 4(-6a) + 2(9a) + (1 - 4a) = 3 \implies a = -1$$

$$\therefore \, a = -1, \, b = 6, \, c = -9, \, d = 5$$

Hence, we get $f(x) = -x^3 + 6x^2 - 9x + 5$.

Answer 2

As $(2,3)$ is the inflection point we have $$f(x)=a(x-2)^3+b(x-2)+3.$$ From $(1,1)$ we get $-a-b+1=1$. Now from $f'(x)=3a(x-2)^2+b$ we get $3a+b=0$, that is $$f(x)=-(x-2)^3+3(x-2)+3.$$