# Question #03f69

Mar 29, 2017

${x}^{3} - 3 {x}^{2} + 9$

#### Explanation:

Let $f \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d$.

Firstly, note that the leading coefficient is 1, which means $a = 1$. So $f \left(x\right) = {x}^{3} + b {x}^{2} + c x + d$.

Now, calculate the derivatives of $f$. We get $f ' \left(x\right) = 3 {x}^{2} + 2 b x + c$, and $f ' ' \left(x\right) = 6 x + 2 b$.

Secondly, since there is an inflection point at $x = 1$, we know that $f ' ' \left(1\right) = 0$, which implies $6 + 2 b = 0$, or $b = - 3$. So $f \left(x\right) = {x}^{3} - 3 {x}^{2} + c x + d$, and $f ' \left(x\right) = 3 {x}^{2} - 6 x + c$.

Thridly, since there is an critical point at $x = 2$, we know that $f ' \left(2\right) = 0$, which implies $3 \times {2}^{2} - 6 \times 2 + c = 0$, or $c = 0$. So $f \left(x\right) = {x}^{3} - 3 {x}^{2} + d$.

Finally, we use the fact that the curve passes through $\left(1 , 7\right)$, so $f \left(1\right) = 7$. Substituting, we get $1 - 3 + d = 7$, so $d = 9$. So $f \left(x\right) = {x}^{3} - 3 {x}^{2} + 9$.