#### To determine

**To find:** The values of
f′−(4) and
f′+(4).

#### Answer

The values of
f′−(4) and
f′+(4) are
−1 and
1 respectively.

#### Explanation

**Result Used:** The left-hand and right-hand derivative is of
f at
x=a

are given by
f′−(a)=limh→0−f(a+h)−f(a)h and
f′+(a)=limh→0+f(a+h)−f(a)h.

**Given:**

The function
f(x)={0if x≤05−xif 0<x<415−xif x≥4

**Calculation:**

Calculate the left-hand derivative of
f at
x=4.

f′−(4) =limh→0−f(4+h)−f(4)h

Since
h<0,
f(4+h)=5−(4+h).

f′−(4) =limh→0−5−(4+h)−15−4h =limh→0−5−4−h−1h =limh→0−(−hh) =−1

Thus, the value of
f−(4) is
−1 .

Calculate the right-hand derivative of
f at
x=4.

f′+(4) =limh→0+f(4+h)−f(4)h

Since
h>0,
f(4+h)=15−(4+h)

f′+(4) =limh→0+15−(4+h)−1h =limh→0+1−{5−(4+h)}5−(4+h)h

Simplify the term in numerator,

f′+(4) =limh→0+1−{1−h}1−hh =limh→0+h1−hh =limh→0+11−h =1

Thus, the value of
f+(4) is 1.

#### To determine

**To sketch:** The graph of
f(x)={0if x≤05−xif 0<x<415−xif x≥4

#### Explanation

**Graph:**

Use the online graphing calculator and draw the graph of
f(x), as shown below in Figure 1.

From the Figure 1, it is observed that the straight line
x=5 is a vertical asymptote to the graph of
f.

#### To determine

**To find:** The points at which
f is discontinuous.

#### Answer

The points at which
f is discontinuous are 0 and 5

#### Explanation

**Result used:** A function
f(x) is continuous at
x=a if
limx→af(x)=f(a).

**Given:**

The function
f(x)={0if x≤05−xif 0<x<415−xif x≥4.

**Calculation:**

**Result used:** A function
f(x) is continuous at
x=a if
limx→af(x)=f(a).

**Given:**

The function
f(x)={0if x≤05−xif 0<x<415−xif x≥4

**Calculation:**

Continuity of
f at
x=0.

The function
f(0)=0 is defined.

The left-hand limit of
f at
x=0 is calculated as follows:

limx→0+f(x)=0

The right-hand limit of
f at
x=0 is calculated as follows.

limx→0+f(x)=5−0 =5

So,
limx→0+f(x)≠limx→0−f(x).

Hence
f is discontinuous at
x=0.

Continuity of
f at
x=4.

The function
f(4)=1 is defined.

The Left-hand limit of
f at
x=4 is calculated as follows:

limx→4+f(x)=5−4 =1

The Right-hand limit of
f at
x=4 is calculated as follows

limx→4+f(x)=15−4 =1

Since,
limx→4+f(x)=limx→4−f(x), the limit exists and equal to 1.

Therefore,
limx→4f(x)=f(4).

Thus,
f is continuous at
x=0.

Continuity of
f at
x=5

The function
f(5)=10 is undefined.

Thus,
f is discontinuous at
x=5

#### To determine

**To find:** The points at which *f* is not differentiable.

#### Answer

The points at which *f* is not differentiable are 0, 4 and 5

#### Explanation

**Given:**

The function
f(x)={0if x≤05−xif 0<x<415−xif x≥4

**Calculation:**

**Differentiability of**
f** at**
x=4.

From part (a), the values of
f′−(4) and
f′+(4) are
−1 and
1 respectively

Hence
f′−(4)≠f′+(4)

Thus,
f is not differentiable at
x=4.

**Differentiability of**
f** at**
x=0,5.

From part (c), the points at which
f is discontinuous are 0 and 5.

Since,
f is discontinuous at the points 0 and 5,
f is not differentiable at 0 and 5.

Thus,
f is not differentiable at the points
x=0,5.