#### To determine

**To find:** The value of g′(3).

#### Answer

The value g′(3) is g′(3)−26.

#### Explanation

**Given:**

The function is g(x)=f(x).

**Result used: Chain Rule**

If *h* is differentiable at *x* and *g* is differentiable at h(x), then the composite function F=g∘h defined by F(x)=g(h(x)) is differentiable at *x* and F′ is given by the product,

F′(x)=g′(h(x))⋅h′(x) (1)

**Formula used:**

The slope of the line passing through the points (x1,y1) and (x2,y2) is y2−y1x2−x1.

**Calculation:**

Obtain the derivative of g(x)=f(x)*.*

g′(x)=ddx(g(x))=ddx(f(x))

Let h(x)=f(x) and g(u)=u where u=h(x).

Apply the chain rule as shown in equation (1),

g′(x)=g′(h(x))⋅h′(x) (2)

The derivative of g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(u)=ddu(u)12

Simplify further and obtain the derivative,

g′(h(x))=12(u)12−1=12(u)1−22=12(u)−12=12u12

Substitute u=f(x) in the above equation,

g′(h(x))=12(f(x))12=12f(x)

Thus, the derivative is g′(h(x))=12f(x).

The derivative of h(x) is computed as follows,

h′(x)=ddx(f(x))=f′(x)

Thus, the derivative is h′(x)=f′(x).

Substitute 12f(x) for g′(h(x)) and f′(x) for h′(x) in equation (2),

F′(x)=12f(x)⋅f′(x)=f′(x)2f(x)

Substitute x=3 in the above equation,

F′(3)=f′(3)2f(3)

From the given graph observe that,f(3)=2.

F′(3)=f′(3)22 (3)

Obtain the slope of f′(3).

From the given graph, it is noticed that the tangent line passing through the points (3,2) and (0,4).

Use the slope formula stated above and compute the slope of the tangent line passing through points (3,2) to (0,4).

f′(3)=4−20−3=−23

Substitute −23 for f′(3) in equation (3),

F′(3)=−2322=−22(3)2=−132

Multiple and divide by 2,

F′(3)=−13222=−23(2)=−26

Therefore, the value g′(3) is g′(3)=−26.