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skelet666 [1.2K]
3 years ago
7

What is the range of the function y=e^4x? y<0 y>0 y<4 y>4

Mathematics
2 answers:
Oliga [24]3 years ago
7 0

We have been given the function y=e^{4x}

We know that the range is set of y values for which the function is defined. Therefore, we will find the value for x and then observe the restriction is y's values.

y=e^{4x}\\\ln y = \ln (e^{4x})\\\ln y =4x\\x=\frac{1}{4}\ln y

Now, we know that logarithm function is not defined for negative values. Hence, the value for y is always greater than zero.

Therefore, the range of the function is given by y>0

B is the correct option.

nirvana33 [79]3 years ago
7 0
This exponential function is increasing .

Whatever the value of x, y is always positive. So the range of y is all value greater than 0==> y>0
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I'm assuming it is asking for domain and range and if it is a function or not


ok, so we know this is quadratic, and the form is a function

domain is all real numbers because no value of x will make a restriction like division by 0 or square roots of a negative

range is from the vertex to either positie or negative infinity
find the vertex
ok, so for y=ax^2+bx+c, the x value of the vertex is -b/2a
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y=1x^2-4x-5
the x value of the vertex is -(-4)/(2*1)=4/2=2
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y=2^2-4(2)-5
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y=-9
and since the coefinet of the x^2 term is positive, it opens up and the vertex is a min
the range is from y=-9 to inifnity


domain=all real numbers
range is y≥-9
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