To determine

To compute: The value Δy and dy for the given values of x and dx=Δx. Sketch the line segments with lengths dy,Δy and dx.

Answer

The values are dy=1 and Δy=1.25.

Explanation

Given:

The function is y=x2−4x. The value x=3 and Δx=0.5

Calculation:

Obtain the value of Δy.

Consider the function f(x)=x2−4x.

Δy=f(x+Δx)−f(x)

Substitute the value x=3 and Δx=0.5,

Δy=f(3+0.5)−f(3)=f(3.5)−f(3)=[(3.5)2−4(3.5)]−[32−4(3)]=[12.5−14]−[9−12]

Simplify the terms,

Δy=−1.75−(−3)=−1.75+3=1.25

Thus, the value Δy=1.25.

Obtain the value of dy.

Note that, the differential is dy=f′(x)dx.

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

f′(x)=ddx(x2−4x)=ddx(x2)−4ddx(x)=2x−4

Substitute 2x−4 of f′(x) in dy=f′(x)dx,

dy=(2x−4)dx

Substitute x=3 and dx=0.5,

dy=(2(3)−4)(0.5)=2(0.5)=1

Thus, the value dy=1.

Therefore, the value dy=1 and Δy=1.25.

Graph:

The graph of y=x2−4x and the line segments with lengths dy,Δy and dx as shown below in Figure 1.

From Figure 1, it is observed that the value of dy and Δy of the function y=x2−4x at x=3 and Δx=0.5 is 1 and 1.25.