Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 259: 23

$\Delta y = 1.25$ $dy = 1$ $dx = \Delta x = 0.5$ We can see a sketch below.

$y = x^2-4x$ $dx = \Delta x = 0.5$ When $x = 3$, then $y = (3)^2-4(3) = -3$ When $x = 3.5$, then $y = (3.5)^2-4(3.5) = -1.75$ $\Delta y = -1.75-(-3) = 1.25$ We can find the value of $dy$: $\frac{dy}{dx} = 2x-4$ $dy = (2x-4)~dx$ $dy = [2(3)-4]~(0.5)$ $dy = 1$ We can see a sketch below.