Question

5. Given the function

(cxl, x < 2
f(x) =
CX - 3, x > 2
Determine the value of c so that the function will be continuous everywhere.

I need #5 with an explanation please

Answer 1

The function

`f(x)=\begin{cases}cx^2&\text{for }x\le2\\cx-3&\text{for }x>2\end{cases}`

is piecewise continuous, since both cx ² and cx - 3 are polynomials. f(x) itself is continuous if both pieces meet at the same defined point. In other words, the limits of f(x) as x → 2 from either side have the same value of f (2) = c•2² = 4c.

We have

`\displaystyle\lim_(x\to2^-)f(x)=\lim_(x\to2)cx^2=4c`

`\displaystyle\lim_(x\to2^+)f(x)=\lim_(x\to2)(cx-3)=2c-3`

so in order for f to be continuous, we need

4c = 2c - 3 → 2c = -3 → c = -3/2