Question

The temperature​ T, in degrees​ Fahrenheit, of a person during an illness is given by the function ​T(t)equalsStartFraction 6 t Over t squared plus 1 EndFraction plus 98.6 ​, where tequals​time, in hours. Find the interval over which the temperature was over 100degrees.

Answer 1

Answer:

't' lies in the interval (0.2476, 4.038)

Explanation:

Given:

The function for temperature (T) as:

$T=\frac{6t}{t^2+1}+98.6$

Now, according to the condition given in the question

$(\frac{6t}{t^2+1}+98.6)>100$

or

$\frac{6t}{t^2+1}>1.4$

or

${6t}>({t^2+1})1.4$

or

6t>(1.4t² + 1.4)

or

(1.4t² + 1.4 - 6t) < 0

on solving the quadratic

we get the intervals as:

t = 0.2476 and t = 4.038

Therefore, the 't' lies in the interval (0.2476, 4.038)

14t^2-80t+28<0  

7t^2-40t+14<0  

Solve by quadratic:  

t=[ 40+/-sqr(1600-4*7*14)]/14  

=.375 or 5.340  

These are the boundaries.  

So the temperature is over 100 in the interval (.375, 5.340) hrs.