# Math Help - Piecewise function

1. ## Piecewise function

For the function f(x) = { -3x + 2 if x < 1
{ x^2 if x > 1

a) sketch the piecewise function

b) determine the x-values, if any, at which the function is discontinuous. Find appropriate limits to support your conclusion.

Attached is my sketch, I don't know if it's correct?
I don't know how to do b.

For the function f(x) = { -3x + 2 if x < 1
{ x^2 if x > 1

a) sketch the piecewise function

b) determine the x-values, if any, at which the function is discontinuous. Find appropriate limits to support your conclusion.

Attached is my sketch, I don't know if it's correct?
I don't know how to do b.

The given function is: $f(x)=\left\{\begin{array}{rcl}-3x+2&if&x<1 \\ x^2&if&x\geq 1\end{array}\right.$

1. The first term belongs to a linear function which is continuous. The 2nd term belongs to a quadratic function which is continuous. The only possible value at which discontinuity could occur is x = 1.

2. Determine

$\lim_{x \to 1 \wedge x<1}(-3x+2)$

and

$\lim_{x \to 1 \wedge x\geq1}(x^2)$

Your sketch is wrong . . .

For the function: . $f(x) \:=\:\begin{Bmatrix}-3x + 2 && x < 1 \\ x^2 && x \ge 1 \end{array}$

a) Sketch the piecewise function

b) Determine the x-values, if any, at which the function is discontinuous.
Find appropriate limits to support your conclusion.
Code:
        *     |          *
*    |
*   |         *
*  |        *
* |      *
*|   *
*
|*  1
- - - - + * + - - - -
|  *
|   o
|

Obviously there is a discontinuity at $x = 1.$

We verify this by taking the limit of $f(x)$ "from both sides".

Take limit of $f(x)$ as $\,x$ approaches 1 "from below".

. . $\displaystyle \lim_{x\to1^-}f(x) \:=\:\lim_{x\to1^-}(-3x+2) \;=\;-1$

Take the limit of $f(x)$ as $\,x$ approaches 1 "from above".

. . $\displaystyle \lim_{x\to1^+} f(x) \;=\;\lim_{x\to1^+}x^2 \;=\;1$

Since the two limits are not equal, $f(x)$ is discontinuous at $x = 1.$

Edit: Too slow again . . . redundant explanation.
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