#### To determine

**To find**: The derivative of the function
f(x) at
x=a if
a≠0.

#### Answer

The value of
f′(a) is
13a−23.

#### Explanation

**Given**:

The given function is,
f(x)=x3.

**Result Used**:

The derivative of a function *f* at
x=a*,* denoted by
f′(x), is

f′(a)=limx→af(x)−f(a)x−a (1)

Difference of cube formula:
x3−y3=(x−a)(x2+y2+xy).

**Calculation**:

Obtain the derivative of the function
f(x) at
x=a.

Compute
f′(a) by using the equation (1),

f′(a)=limx→af(x)−f(a)x−a=limx→ax3−a3x−a=limx→a(x13−a13)x−a=limx→a(x13−a13)x−a×(x23+a13x13+a23)(x23+a13x13+a23)

Apply the difference of cube formula in the numerator as follows,

f′(a)=limx→a(x13)3−(a13)3x−a×1(x23+a13x13+a23)=limx→ax−ax−a×1(x23+a13x13+a23)=limx→a1{x23+a13x13+a23} [∵ x−a≠0]=limx→a(1)limx→a{x23+a13x13+a23}

=1a23+a13a13+a23

Simplify the denominator,

f′(a)=1a23+a23+a23=13a23=13a−23

Thus, the value of
f′(a) is
13a−23.

#### To determine

**To Show:** The function
f′(0) does not exist.

#### Explanation

**Result used:**

The derivative of a function *f ,* denoted by
f′(x), is

f′(a)=limh→0f(a+h)−f(a)h (2)

**Proof:**

Consider the function
f(x)=x3**.**

Compute
f′(0) by using the equation (2),

f′(0)=limh→00+h3−03h=limh→0h3−0h=limh→0h13h=limh→01h⋅h−13

=limh→01h23

Here, the function
1h23 becomes larger and larger as *h* tends to zero. That is,

f′(0)=limh→01h23=∞

Therefore, the derivative of the function does not exist at
x=0.

Thus, the required proof is obtained.

#### To determine

**To show:** The
y=x3 has a vertical tangent line at
(0,0)

#### Explanation

**Result Used:**

A curve has a vertical tangent line at
x=a if *f* is continuous at
x=a and
limx→a|f′(x)|=∞

**Proof:**

Consider the equation
f(x)=x3.

Substitute
x=0 in
f(x),

f(0)=03=0

Thus
f(0)=0 is defined.

The limit of the function
f(x) is computed as follows,

limx→0f(x)=limx→0x3=03=0

Therefore,
f(x) is continuous at
x=0.

From part (a),
f′(x)=x−233

Take the limit of the function
f′(x) as *x* approaches zero.

limx→0|f′(x)|=limx→0|13x23|=13limx→0|1x23|=13limx→01x23=∞ [From part (b)]

Since the function
f(x) is continuous at
x=0 and
limx→0|f′(x)|=∞.

By result, the curve
f(x) has a vertical tangent at
x=0.

Thus, the curve
f(x) has a vertical tangent at the point
(0,0).

**Graph:**

Use the online graphing calculator to draw the graph of the function
y=x3 as shown below in Figure 1,

From Figure 1, it is clear that the *y*-axis is the vertical tangent to the curve
y=x3.