Answer
(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 $0.1$
Work Step by Step
(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 $0.1$
