Chapter 3.5, Problem 1E

a.

To determine

To Calculate: The general formulas for $\Delta y\&dy$

Answer to Problem 1E

  $\Delta y=\left(4x-4\right)\left(\Delta x\right)+2{\left(\Delta x\right)}^2,dy=\left(4x-4\right)dx$ .

Explanation of Solution

Given information:

  $y=2x^2-4x+5$

Concept and Formula Used:

  $\Delta y=f(x+\Delta x)-f(x),dy=f\text{'}(x)\Delta x=f\text{'}(x)dx$

Calculation:

The equation given is

  $\begin{array}{l}f(x)=y=2x^2-4x+5\\ \Delta y=f(x+\Delta x)-f(x)\\ $ \Delta y=2{(x+\Delta x)}^2-4(x+\Delta x)+5-\left(2x^2-4x+5\right)$\Delta y=2\left(x^2+2x\left(\Delta x\right)+{\left(\Delta x\right)}^2\right)-4(x+\Delta x)+5-\left(2x^2-4x+5\right)\\ \Delta y=2x^2+4x\left(\Delta x\right)+2{\left(\Delta x\right)}^2-4x-4\left(\Delta x\right)+5-2x^2+4x-5\\ \Delta y=4x\left(\Delta x\right)+2{\left(\Delta x\right)}^2-4\left(\Delta x\right)\\ \Delta y=\left(4x-4\right)\left(\Delta x\right)+2{\left(\Delta x\right)}^2\end{array}$

Now,

  $\begin{array}{l}dy=f\text{'}(x)dx\\ dy=\frac{d}{dx}\left(2x^2-4x+5\right)dx\\ dy=\left(4x-4\right)dx\end{array}$

Conclusion:

  $\Delta y=\left(4x-4\right)\left(\Delta x\right)+2{\left(\Delta x\right)}^2,dy=\left(4x-4\right)dx$

b.

To determine

To Calculate: The value of $\Delta y\&dy$ when ‘x’ changes from $a\text{ to }a+\Delta x$ where

  $a=2\&\Delta x=-0.2$

Answer to Problem 1E

  $\Delta y=-0.72,dy=-0.8$ .

Explanation of Solution

Concept and Formula Used:

  $\Delta y=f(x+\Delta x)-f(x),dy=f\text{'}(x)\Delta x=f\text{'}(x)dx$

Calculation:

Igt is given that x changes from $a\text{ to }a+\Delta x$ i.e. from $x=2\&x=2-0.2=1.8$

The equation given is

  $\begin{array}{l}f(x)=y=2x^2-4x+5\\ \Delta y=f(x+\Delta x)-f(x)\\ $ \Delta y=2{(x+\Delta x)}^2-4(x+\Delta x)+5-\left(2x^2-4x+5\right)$\Delta y=2\left(x^2+2x\left(\Delta x\right)+{\left(\Delta x\right)}^2\right)-4(x+\Delta x)+5-\left(2x^2-4x+5\right)\\ \Delta y=2x^2+4x\left(\Delta x\right)+2{\left(\Delta x\right)}^2-4x-4\left(\Delta x\right)+5-2x^2+4x-5\\ \Delta y=4x\left(\Delta x\right)+2{\left(\Delta x\right)}^2-4\left(\Delta x\right)\\ \Delta y=\left(4x-4\right)\left(\Delta x\right)+2{\left(\Delta x\right)}^2\end{array}$

Putting

  $\begin{array}{l}x=2\&\Delta x=-0.2\\ \Delta y=\left(4\left(2\right)-4\right)\left(-0.2\right)+2{\left(-0.2\right)}^2\\ \Delta y=-0.8+0.08\\ \Delta y=-0.72\end{array}$

Now,

  $\begin{array}{l}dy=f\text{'}(x)dx\\ dy=\frac{d}{dx}\left(2x^2-4x+5\right)dx\\ dy=\left(4x-4\right)dx\end{array}$

Putting

  $\begin{array}{l}x=2\&dx=\Delta x=-0.2\\ dy=\left(4\left(2\right)-4\right)\left(-0.2\right)\\ dy=-0.8\end{array}$

Conclusion:

  $\Delta y=-0.72,dy=-0.8$