## Calculus: Early Transcendentals 8th Edition

(a) $33.0~W$ (b) The allowed range of wattage is $32.9 \leq w \leq 33.1$ (c) $w$ is $x$ $(0.1w^2+2.155w+20)$ is $f(x)$ $33.0$ is $a$ $200$ is $L$ The given value of $\epsilon$ is $1$ The corresponding value of $\delta$ is $0.1$
(a) $T(w) = 0.1w^2+2.155w+20$ We can find $w$ when $T = 200^{\circ}C$: $T(w) = 0.1w^2+2.155w+20 = 200$ $0.1w^2+2.155w-180 = 0$ We can use the quadratic formula to find the required power $w$: $w = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $w = \frac{-2.155\pm \sqrt{(2.155)^2-(4)(0.1)(-180)}}{(2)(0.1)}$ $w = \frac{-2.155\pm \sqrt{76.644025}}{0.2}$ $w = -54.6, 33.0$ Since power should be positive, the required power is $33.0~W$ (b) We can find $w$ when $T = 199^{\circ}C$: $T(w) = 0.1w^2+2.155w+20 = 199$ $0.1w^2+2.155w-179 = 0$ We can use the quadratic formula to find the required power $w$: $w = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $w = \frac{-2.155\pm \sqrt{(2.155)^2-(4)(0.1)(-179)}}{(2)(0.1)}$ $w = \frac{-2.155\pm \sqrt{76.244025}}{0.2}$ $w = -54.4, 32.9$ We can find $w$ when $T = 201^{\circ}C$: $T(w) = 0.1w^2+2.155w+20 = 201$ $0.1w^2+2.155w-181 = 0$ We can use the quadratic formula to find the required power $w$: $w = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $w = \frac{-2.155\pm \sqrt{(2.155)^2-(4)(0.1)(-181)}}{(2)(0.1)}$ $w = \frac{-2.155\pm \sqrt{77.044025}}{0.2}$ $w = -54.7, 33.1$ The allowed range of wattage is $32.9 \leq w \leq 33.1$ (c) $\lim\limits_{x \to a}f(x) = L$ $\lim\limits_{w \to 33.0}(0.1w^2+2.155w+20) = 200$ $w$ is $x$ $(0.1w^2+2.155w+20)$ is $f(x)$ $33.0$ is $a$ $200$ is $L$ The given value of $\epsilon$ is $1$ The corresponding value of $\delta$ is $0.1$