Question

If $f\left(x\right)=4x^3-3x^2+2x+k$ and $f\left(0\right)=1,f\left(1\right)=4$, find $f\left(x\right)$.

Answer 1

It is given that $f\left(0\right)=1$ and $f\left(1\right)=4$ for the polynomial $f\left(x\right)=4x^3-3x^2+2x+k$, therefore, we substitute the values and find the value of $k$ as shown below:

$Whenf\left(0\right)=1f\left(0\right)=4(0{)}^3-3(0{)}^2+2\left(0\right)+k\Rightarrow 1=kWhenf\left(1\right)=4f\left(1\right)=4(1{)}^3-3(1{)}^2+2\left(1\right)+k\Rightarrow 4=4-3+2+k\Rightarrow 4=3+k\Rightarrow k=4-3\Rightarrow k=1$

We conclude that, in both the cases $k=1$, thus, we substitute $k=1$ in $f\left(x\right)$:

$f\left(x\right)=4x^3-3x^2+2x+1$

Hence, $f\left(x\right)=4x^3-3x^2+2x+1$.