Trying to solve problem related to continuous functions

Question

A function $f$ is defined by $$ \begin{equation} f(x)= \begin{cases} -2x^3-3 \text{ for } x < -1 \\ ax+b \text{ for } -1 \leq x < 1 \\ 2x^2+1 \text{ for } 1 \leq x\\ \end{cases} \end{equation} $$

where $a$ and $b$ are constants. The function $f$ is continuous. Find $a$ and $b$

Answer 1

Compute $−2x^3−3$ at $x=−1$ which must be equal to $ax+b$ at $x=−1$.

Similarly, compute $2x^2+1$ which must coincide with $ax+b$ at $x=1$.

From 1 and 2 cases we get $a-b=1$

From 2 and 3 cases we get $a+b=3$

Thus $2a=4$, hence $a=2$, therefore $b=1$

Answer 2

Easy solution: Use the fact that continuous univariate functions when graphed have no jumps. So compute $−2x^3−3$ at $x=-1$ which must be equal to $ax+b$ at $x=-1$. Similarly, $2x^2+1$ must coincide with $ax+b$ at $x=1$. Then you just have a system of equations of two variables.