Question

23-26: Compute \(\Delta y\) and \(dy\) for the given values of x and \(dx = \Delta x\). Then sketch a diagram like a Figure 5 showing the line segments with lengths \(dx\), \(dy\), and \(\Delta y\).

23. \(y = {x^2} - 4x,{\rm{ }}x = 3,{\rm{ }}\Delta x = 0.5\)

Short answer

The values of \(\Delta y\) and \(dy\)are 1.25 and 1.

Step-by-step solution

Definition of differentials

The equation establishes the differential \(dy\)with respect to \(dx\) as shown below:

\(dy = f'\left( x \right)dx\)

Where \(dx\) represent the independent variableand \(dy\) represent the dependent variable.

Consider \(P\left( {x,f\left( x \right)} \right)\) and \(Q\left( {x + \Delta x,f\left( {x + \Delta x} \right)} \right)\) as the points on the graph of f and consider \(dx = \Delta x\). The change in \(y\) is given by \(\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)\).

Compute the values of \(\Delta y\) and \(dy\)

The given function is \(y = f\left( x \right) = {x^2} - 4x\).

Substitute \(x = 3\) and \(\Delta x = 0.5\) to obtain the values of \(\Delta y\)as shown below:

\(\begin{aligned}\Delta y &= f\left( {3 + 0.5} \right) - f\left( 3 \right)\\ &= f\left( {3.5} \right) - f\left( 3 \right)\\ &= {\left( {3.5} \right)^2} - 4\left( {3.5} \right) - \left( {{{\left( 3 \right)}^2} - 4\left( 3 \right)} \right)\\ &= - 1.75 - \left( { - 3} \right)\\ &= 1.25\end{aligned}\)

The value of\(\Delta y\)is 1.25.

Obtain the differential of the function as shown below:

\(\begin{aligned}dy &= f'\left( x \right)dx\\dy &= \left( {2x - 4} \right)dx\end{aligned}\)

Substitute\(x = 3\)and \(dx = \Delta x = 0.5\)to obtain\(dy\)as shown below:

\(\begin{aligned}dy &= \left( {2\left( 3 \right) - 4} \right)\left( {0.5} \right)\\ &= \left( {6 - 4} \right)\left( {0.5} \right)\\ &= 2\left( {0.5} \right)\\ &= 1\end{aligned}\)

Thus, the value of \(dy\) is 1.

Sketch the graph to show the line segments

The function is \(y = {x^2} - 4x\) and \(\Delta x = 0.5\). There is a tangent line to the function at\(x = 3\). The values of \(\Delta y\) and \(dy\)are 1.25 and 1.

Use the above points to sketch the geometric figure of the function as shown below:

Thus, the graphs show the line segment with lengths \(dx,\)\(\Delta y\) and \(dy\).