Respuesta :
The given function is [tex]y=x^2+7[/tex].
Minimum or maximum value:
At the extremum (maximum or minimum) value, the function will have zero slope. So, differentiate the given function once and equate it to zero to get the extremum point.
dy/dx=0
[tex]\Rightarrow 18x=0\cdots(i)[/tex]
[tex]\Rightarrow x=0[/tex]
Now, check whether the point x=0 is corresponding to the maximum value or minimum value by differentiating the function twice,
[tex]\frac {d^2y}{dx^2}=18[/tex]
As [tex]\frac {d^2y}{dx^2} >0[/tex] for all value of x, so x=0 is the point corresponding to minima.
Put x=0 in the given function to get the minimum value.
[tex]y_{min}=9(0^2)+7[/tex]
[tex]y_{min}=7[/tex]
Domain and range:
The function defined for all the values of the independent variable, x.
So, the domain is [tex](-\infty, \infty)[/tex].
The range of the function is the possible value of y.
The minimum value, for x=0, is y=7.
The maximum value, as [tex]x\rightarrow \infty \;or\; -\infty, y\rightarrow \infty[/tex].
Hence the range of the function is [tex][7,\infty)[/tex].
The value of x for which the function is increasing and decreasing:
If the slope of the function is negative than the function is decreasing, so
Then, from equation (i), the value of x for which dy/dx<0,
18x<0
[tex]\Rightarrow x<0[/tex]
Hence, the function is decreasing for [tex]x\in {-\infty, 0)[/tex] .
While if the slope of the function is positive than the function is increasing, so
Then, from equation (i), the value of x for which dy/dx<0,
18x>0
[tex]\Rightarrow x>0[/tex]
Hence, the function is increasing for [tex]x\in {0,\infty)[/tex]
