Answer:
See below
Step-by-step explanation:
Extreme values of a function are found by taking the first derivative of the function and setting it equal to 0. To determine if it's a minimum or maximum, we set the second derivative equal to 0 and determine if its positive or negative respectively.
Let's do [tex]f(x)=3x^4+2x^3-5x^2+7[/tex] as an example
By using the power rule where [tex]\frac{d}{dx}(x^n)=nx^{n-1}[/tex], then [tex]f'(x)=12x^3+6x^2-10x[/tex]
Now set [tex]f'(x)=0[/tex] and solve for [tex]x[/tex]:
[tex]0=12x^3+6x^2-10x[/tex]
[tex]0=2x(6x^2+3x-5)[/tex]
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-3\pm\sqrt{3^2-4(6)(-5)}}{2(6)}[/tex]
[tex]x=\frac{-3\pm\sqrt{9+120}}{12}[/tex]
[tex]x=\frac{-3\pm\sqrt{129}}{12}[/tex]
[tex]x=-\frac{3}{12}\pm\frac{\sqrt{129}}{12}[/tex]
[tex]x=-\frac{1}{4}\pm\frac{\sqrt{129}}{12}[/tex]
[tex]x_1=0,x_2\approx0.6965,x_3=-1.1965[/tex]
By plugging our critical points into [tex]f(x)[/tex], we can see that our extreme values are located at [tex](0,7)[/tex], [tex](0.6965,5.956)[/tex], and [tex](-1.1965,2.565)[/tex].
The second derivative would be [tex]f''(x)=36x^2+12x-10[/tex] and plugging in our critical points will tell us if they are minimums or maximums.
If [tex]f''(x)>0[/tex], it's a minimum, but if [tex]f''(x)<0[/tex], it's a maximum.
Since [tex]f''(0)=-10<0[/tex] then [tex](0,7)[/tex] is a local maximum
Since [tex]f''(0.6965)=15.822>0[/tex], then [tex](0.6965,5.956)[/tex] is a local minimum
Since [tex]f''(-1.1965)=27.18>0[/tex], then [tex](-1.1965,2.565)[/tex] is a global minimum
Therefore, the extreme values of [tex]f(x)=3x^4+2x^3-5x^2+7[/tex] are a global minimum of [tex](-1.1965,2.565)[/tex], a local minimum of [tex](0.6965,5.956)[/tex], and a local maximum of [tex](0,7)[/tex].
Hope this example helped you understand! I've attached a graph to help you visualize the extreme values and where they're located.
