Respuesta :
The value of the function f(x)=6/-2+2[tex]e^{-0.3x}[/tex] at x=4 after rounded to the nearest tenth is -2.4.
Given Function of x : f(x)=6/-2+2[tex]e^{-0.3x}[/tex]
The given function showing the relationship between x and f(x) is f(x)=6/-2+2[tex]e^{-0.3x}[/tex] where e is exponential and we have to find the value of f(4) by just putting the value of x=4 in the given function and round to the nearest tenth means replacing a number with an approximate value.
f(4)=6/-2+2[tex]e^{-0.3*4}[/tex] ( The value of e is 2.7812)
f(4)=-3+2[tex]e^{-1.2}[/tex]
f(4)=-3+2*0.30119 (The value of [tex]e^{-1.2}[/tex] is 0.30119)
f(4)=-3+0.60238
f(4)=-2.3972
Round -2.3972 to tenth is -2.4.
Hence the value of function of at x=4 is -2.4.
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The value of f(4) is -4.3.
The function is a relation between two different sets X and set Y which are in one-one or many-one relation. Here set X is called as the domain and set Y is called as the codomain.
For example: f(x)=3x+9
To get the value of the function at any point of x, we have to put the number in the function to get its function value.
For example the value of f(x) in above function at x=a will be f(a)=3a+9
Similarly given function in the question
[tex]f(x)=\frac{6}{-2+2e^{-0.3x}}[/tex]
to get the value of f(x) at x=4 put the value of x in the function,
[tex]f(x)=\frac{6}{-2+2e^{-0.3*4}}[/tex]
⇒[tex]f(x)=\frac{6}{-2+2e^{-1.2}}[/tex]
⇒f(x)=-4.3
Therefore the value of f(4) is -4.3.
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