Question

The function $t$ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t\left(C\right)=\frac{9C}{5}+{32}^o$
Find :

(i) $t\left(0^o\right)$

(ii) $t\left({28}^o\right)$

(iii) $t(-{10}^o)$

(iv) The value of $C$ when $t\left(C\right)={212}^{o}F$

Answer 1

The given function is

$t\left(C\right)=\frac{9C}{5}+{32}^o$
(i) $t\left(0^o\right)=\frac{9\times 0}{5}+{32}^o=0+{32}^o={32}^{o}F$

(ii) $t\left({28}^o\right)=\frac{9\times {28}^o}{5}+{32}^o=\frac{{252}^o+{160}^o}{5}=\frac{{412}^o}{5}={82.4}^{o}F$

(iii) $t(-{10}^o)=\frac{9\times (-{10}^o)}{5}+{32}^o=9\times (-2^o)+{32}^o=-{18}^o+{32}^o={14}^{o}F$

(iv) It is given that $t\left(C\right)={212}^{o}F$

$\therefore {212}^o=\frac{9C}{5}+{32}^o$

$\Rightarrow \frac{9C}{5}={212}^o-{32}^o$

$\Rightarrow \frac{9C}{5}={180}^o$

$\Rightarrow 9C={180}^o\times 5$

$\Rightarrow C=\frac{{180}^o\times 5}{9}={100}^o$

Thus the value of Celsius temperature is ${100}^o$ when Fahrenheit temperature is ${212}^o$.