The thermic effect of food (TEF) is the increase in metabolic rate after a meal. To model TEF which is measured in kJ/h, researchers use the following functions: S(t) = 175.9teiš 9 (t) = 113.6teT where ƒ (t) is the TEF for a lean person and g(t) is the TEF for an obese person respectively and t, is the time in hours. Find the marimum value of the TEF for both individuals. Solve the above, by answering the following questions: a) Examine both functions above and then rewrite the TEF as a single generalfunction h(t) by replacing the numerical constants in f (t) and g (t) with the symbolic constants a and b. (b) Calculate the derivative h' (t). (c) Find the intervals of increase or decrease of h(t). (d) Using your answerfound in part (c) above, find the maximum TEF for a lean person and the maximum TEF for an obese person respectively by substituting the appropriate valuesfor a and b. (e) Calculate the derivative h" (t). (f) Using your answerfound in part (e) above, find the inflection point for the graph of f (t) (for a lean person) and for the graph of g (t) (for an obese person) respectively and interpret in your own words the physical meaning of this term.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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I just need help with 1d) e) f) ,I have attached answers a b and c for refrence

The thermic effect of food (TEF) is the increase in metabolic rate after a meal. To model TEF which
is measured in kJ/h, researchers use the following functions:
S (t) = 175.9teīš
g (t) = 113.6te
where f (t) is the TEF for a lean person and g (t) is the TEF for an obese person respectively and t,
is the time in hours. Find the marimum value of the TEF for both individuals.
Solve the above, by answering the following questions:
a) Examine both functions above and then rewrite the TEF as a single generalfunction h(t) by
replacing the numerical constants in f (t) and g (t) with the symbolic constants a and b.
(b) Calculate the derivative h' (t).
(c) Find the intervals of increase or decrease of h(t).
(d) Using your answerfound in part (c) above, find the maximum TEF fora lean person and the
maximum TEF for an obese person respectively by substituting the appropriate values fora and b.
(e) Calculate the derivative h" (t).
(f) Using your answerfound in part (e) above, find the inflection point for the graph of f (t) (for a lean
person) and for the graph of g (t) (for an obese person) respectively and interpret in your own words
the physical meaning of this term.|
Transcribed Image Text:The thermic effect of food (TEF) is the increase in metabolic rate after a meal. To model TEF which is measured in kJ/h, researchers use the following functions: S (t) = 175.9teīš g (t) = 113.6te where f (t) is the TEF for a lean person and g (t) is the TEF for an obese person respectively and t, is the time in hours. Find the marimum value of the TEF for both individuals. Solve the above, by answering the following questions: a) Examine both functions above and then rewrite the TEF as a single generalfunction h(t) by replacing the numerical constants in f (t) and g (t) with the symbolic constants a and b. (b) Calculate the derivative h' (t). (c) Find the intervals of increase or decrease of h(t). (d) Using your answerfound in part (c) above, find the maximum TEF fora lean person and the maximum TEF for an obese person respectively by substituting the appropriate values fora and b. (e) Calculate the derivative h" (t). (f) Using your answerfound in part (e) above, find the inflection point for the graph of f (t) (for a lean person) and for the graph of g (t) (for an obese person) respectively and interpret in your own words the physical meaning of this term.|
1.a) f(t) =175,9 te s
= atb-t
%3D
- /13.6 te
= 113:6+(e (te) *
= atb-t
gereral
funchion
is hl4)=atb-/
b) h 't) = d (atb-t)
elt
= ab -t d
dt
9.
= ab -t -atb-1 (Inb)
h= abt (1-tInb)
; derivitive of funchion a: bilE) = abt (1-tHnb)
c) n'lt) =CO
abit (l-Hnb)=0
1-Hnb =0
In.
* The funchion is imcreasing in The interval (-e ; hs
dee rasing in
the
inkral
and
In)
Transcribed Image Text:1.a) f(t) =175,9 te s = atb-t %3D - /13.6 te = 113:6+(e (te) * = atb-t gereral funchion is hl4)=atb-/ b) h 't) = d (atb-t) elt = ab -t d dt 9. = ab -t -atb-1 (Inb) h= abt (1-tInb) ; derivitive of funchion a: bilE) = abt (1-tHnb) c) n'lt) =CO abit (l-Hnb)=0 1-Hnb =0 In. * The funchion is imcreasing in The interval (-e ; hs dee rasing in the inkral and In)
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