For `t in (-infty,3)` function `f` is defined as `f(t)=t^2-c` and since quadratic function is continuous it means the function `f` is continuous on `(-infty,3)` for any constant `c`. Similarly for `t in (3,infty)` function `f` is continuous for any constant `c` because linear function is continuous.

Now we...

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For `t in (-infty,3)` function `f` is defined as `f(t)=t^2-c` and since quadratic function is continuous it means the function `f` is continuous on `(-infty,3)` for any constant `c`. Similarly for `t in (3,infty)` function `f` is continuous for any constant `c` because linear function is continuous.

Now we only need to find such `c` that function is continuous at `t=3`. For that to hold both `t^2-c` and `ct+1` must have the same value at `t=3`. Thus we need to solve equation

`3^2-c=3c+1`

`4c=8`

`c=2`

**Your solution is** `c=2`.

The graph shows function `f(t)={(t^2-2 if x<3),(2t+1 if x>=3):}` which is your function for `c=2` and it is, as you can see, continuous.