max is at vertex
in form [tex]y=ax^2+bx+c[/tex]
the x value of the vertex is [tex]\frac{-b}{2a}[/tex]
given, [tex]y=6x^2-1[/tex]
a=6, b=0
the x value of the vertex is -0/(2*6)=0
the y value is [tex]y=6(0)^2-1=0-1=-1[/tex]
so vertex is at (0,-1)
since the value of a is positive, the parabola opens up and the vertex is a minimum value of the function
therefore that value is the smallest value the function can be
domain=numbesr you can use for x
range=numbesr you get out of inputting the domain
domain=all real numbers
range is {y | y≥-1} since y=-1 is the minimum
The minimum or maximum, domain and range of the given function is required.
Domain: [tex]x\in (-\infty,\infty)[/tex]
Range: [tex]y\in [-1,\infty)[/tex]
The function is minimum at [tex](0,-1)[/tex]
The given function is [tex]y=6x^2-1[/tex]
The domain consists of the [tex]x[/tex] values for which the function is defined.
The function is defined for all values of [tex]x[/tex] so, the domain is
[tex]x\in (-\infty,\infty)[/tex]
The range is the list of [tex]y[/tex] values for all values of [tex]x[/tex].
The range here is [tex]y\in [-1,\infty)[/tex]
Differentiating with respect to [tex]x[/tex]
[tex]y'=12x[/tex]
Equating to 0
[tex]0=12x\\\Rightarrow x=0[/tex]
Double derivative of the function
[tex]y''=12[/tex]
Since, [tex]y''>0[/tex], it will only have a minimum value.
Substituting [tex]x=0[/tex]
[tex]y=6\times 0-1\\\Rightarrow y=-1[/tex]
The function is minimum at [tex](0,-1)[/tex]
The function does not have a maximum point.
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