Piecewise function graph interpretation

Question

Examine the function

f(x) = {x^(2/3), x< 0...(x-1)^2-1, 0 ≤ x ≤ 3.....1/(x-3), x > 3 for points of discontinuity and non-differentiability.

b. Examine the graph of this function. At what points is the function discontinuous? Give reasons for your answers.

c. At what points is the function non-differentiable? Give reasons for your answers.

Note: You may need to examine “suspicious” points more closely by using a close-up graph or some slope tables.

Answer 1

A function is Discontinuous if Limit as X approaches A [Left Side] for f(x) =/= Limit as X approaches A [Right Side] for f(x)

Lim X => 3 [Left Side] for f(x):

1. [(x - 1) ^2] - 1
2. [(3 - 1) ^2] -1
3. 3

Lim X => 3 [Right Side] for f(x):

1. 1 / (x - 3)
2. 1 / (3 - 3)
3. Undefined

Limit as X approaches 3 [Left Side] for f(x) =/= Limit as X approaches 3 [Right Side] for f(x)

Limit as X approaches 3 for f(x) Does not exist => Discontinuity @ X = 3

A function is Non-Differentiable if Limit as X approaches A [Left Side] for f ' (x) =/= Limit as X approaches A [Right Side] for f ' (x)

f ' (x) as X approaches 0 [Left Side] = [2/3] * [x ^ (-1/3)]

f ' (x) as X approaches 0 [Right Side] and X approaches 3 [Left Side] = 2x - 2

f ' (x) as X approaches 3 [Right Side] = -1 / (x^2 - 6x + 9)

Lim X => 0 [Left Side] for f ' (x):

1. [2/3] * [x ^ (-1/3)]
2. [2/3] * [0 ^ (-1/3)]
3. 0

Lim X => 0 [Right Side] for f ' (x):

1. 2x - 2
2. 2(0) - 2
3. -2

Lim X => 3 [Left Side] for f ' (x):

1. 2x - 2
2. 2(3) - 2
3. 4

Lim X => 3 [Right Side] for f ' (x):

1. -1 / (x^2 - 6x + 9)
2. -1 / (3^2 - 6x + 9)
3. Undefined

Limit as X approaches 0 [Left Side] for f ' (x) =/= Limit as X approaches 0 [Right Side] for f ' (x)

Limit as X approaches 0 for f ' (x) Does not exist => Non-Differentiable @ X = 0

Limit as X approaches 3 [Left Side] for f ' (x) =/= Limit as X approaches 3 [Right Side] for f ' (x)

Limit as X approaches 3 for f ' (x) Does not exist => Non-Differentiable @ X = 3