The equation C = 5 9 F − 160 9 gives the relation between temperature readings in Celsius and Fahrenheit.

(a) Is C a function of F? Yes, C is a function of F. No, C is a not a function of F.

(b) What is the mathematical domain of this function? (Enter your answer using interval notation. If C is not a function of F, enter DNE.)

(c) If we consider this equation as relating temperatures of water in its liquid state, what are the domain and range? (Enter your answers using interval notation. If C is not a function of F, enter DNE.) domain range

(d) What is C when F = 71°? (Round your answer to two decimal places. If C is not a function of F, enter DNE.) C(71) = °C

Therefore, giving that we know the value in Fahrenheit ''F'' we can find the reading in Celsius ''C''. This define a function C(F) that depends of the variable ''F''.

So for the incise (a) we answer Yes, C is a function of F.

For (b) we need to find the mathematical domain of this function. Giving that we haven't got any mathematical restriction, the mathematical domain of the function are all real numbers.

Dom (C) = ( - ∞ , + ∞)

For (c) we know that the water in liquid state and at normal atmospheric pressure exists between 0 and 100 Celsius.

Therefore the range will be

Rang (C) = (0,100)

Now, we need to find the domain for this range. We do this by equaliting and finding the value for the variable ''F'' :

For C = 0 :

[tex]0=\frac{5}{9}F-\frac{160}{9}[/tex] ⇒ [tex]F=32[/tex]

And for C = 100 :

[tex]100=\frac{5}{9}F-\frac{160}{9}[/tex] ⇒ [tex]F=212[/tex]

Therefore, the domain as relating temperatures of water in its liquid state is

Dom (C) = (32,212)

For (d) we only need to replace in the equation by [tex]F=71[/tex] and find the value of C ⇒

[tex]C=\frac{5}{9}F-\frac{160}{9}[/tex] ⇒

[tex]C=(\frac{5}{9})(71)-\frac{160}{9}[/tex]

[tex]C=\frac{65}{3}[/tex] ≅ 21.67

C(71) = 21.67 °C

akposevictor akposevictor · Answer 2 · 2022-01-27T16:57:58+01:00

a. Yes, C is a function of F.

b. Domain of the function is: ( - ∞ , + ∞)

c. Range of function C is: (0, 100)

the domain of function C is: (32, 212).

d. [tex]\mathbf{C(71) = 21.67^{\circ}C}[/tex]

Range and Domain of A Function

The domain of a function is the input value, while the range is the corresponding output value.

a. Given the equation of the function as, [tex]C = \frac{5}{9}(F) - \frac{160}{9}[/tex], this means that, the value of C is dependent on the value of F, therefore, the answer is, Yes, C is a function of F.

b. Since no mathematical restriction was stated, therefore:

Domain of the function is: ( - ∞ , + ∞)

c. Temperature of water in its liquid state ranges between 0 and 100°C, therefore:

Range of function C is: (0, 100)

To find the domain, substitute C = 0 and C = 100, respectively, into [tex]C = \frac{5}{9}(F) - \frac{160}{9}[/tex].

Thus:

[tex]0 = \frac{5}{9}(F) - \frac{160}{9}\\\\F = 32[/tex]

[tex]100 = \frac{5}{9}(F) - \frac{160}{9}\\\\F = 212[/tex]

Therefore, the domain of function C is: (32, 212).

d. To find C when F = 71°, substitute F = 71° into [tex]C = \frac{5}{9}(F) - \frac{160}{9}[/tex].

Thus:

[tex]C(71) = \frac{5}{9}(71) - \frac{160}{9}\\\\C(71) = \frac{65}{3} \\\\\mathbf{C(71) = 21.67^{\circ}C}[/tex]

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