Question

How to find extreme values of a function.

Answer 1

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 $f(x)=3x^4+2x^3-5x^2+7$ as an example

By using the power rule where $\frac{d}{dx}(x^n)=nx^{n-1}$, then $f'(x)=12x^3+6x^2-10x$

Now set $f'(x)=0$ and solve for $x$:

$0=12x^3+6x^2-10x$

$0=2x(6x^2+3x-5)$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-3\pm\sqrt{3^2-4(6)(-5)}}{2(6)}$

$x=\frac{-3\pm\sqrt{9+120}}{12}$

$x=\frac{-3\pm\sqrt{129}}{12}$

$x=-\frac{3}{12}\pm\frac{\sqrt{129}}{12}$

$x=-\frac{1}{4}\pm\frac{\sqrt{129}}{12}$

$x_1=0,x_2\approx0.6965,x_3=-1.1965$

By plugging our critical points into $f(x)$, we can see that our extreme values are located at $(0,7)$, $(0.6965,5.956)$, and $(-1.1965,2.565)$.

The second derivative would be $f''(x)=36x^2+12x-10$ and plugging in our critical points will tell us if they are minimums or maximums.

If $f''(x)>0$, it's a minimum, but if $f''(x)<0$, it's a maximum.

Since $f''(0)=-10<0$ then $(0,7)$ is a local maximum

Since $f''(0.6965)=15.822>0$, then $(0.6965,5.956)$ is a local minimum

Since $f''(-1.1965)=27.18>0$, then $(-1.1965,2.565)$ is a global minimum

Therefore, the extreme values of $f(x)=3x^4+2x^3-5x^2+7$ are a global minimum of $(-1.1965,2.565)$, a local minimum of $(0.6965,5.956)$, and a local maximum of $(0,7)$.

Hope this example helped you understand! I've attached a graph to help you visualize the extreme values and where they're located.