Question

Find the local maxima and local minima for the given function and also find the local maximum and local minimum values $g\left(x\right)=x^3-3x$

Answer 1

Maximum or minimum can be seen by using derivatives.

Steps1: First find first derivative of the function

Step2: Put it equal to zero and find $x$ were first derivative is zero

Step3: Now find second derivative

Step4: Put $x$ for which first derivative was zero in equation of second derivative

Step5: If second derivative is greater than zero then function takes minimum value at that $x$ and if second derivative is negative then function will take maximum value at that $x$. If Second derivative is zero them it means that this is the point of inflection.

$f^{{}^{\prime }}\left(x\right)=3x^2-3$

Putting this equal to zero, we get
$3x^2-3=0$
$\Rightarrow x=\pm 1$.

Now let's see the double derivative of this function.
$f^{{}^{\prime \prime }}\left(x\right)=6x$
At $x=1$
$f^{{}^{\prime \prime }}\left(1\right)=6$
Which is positive, this means function will take minimum value at $x=1$
Minimum value of the function is $f\left(1\right)=-2$
At $x=-1$
$f^{{}^{\prime \prime }}(-1)=-6$
Which is negative, this means function will take maximum value at $x=-1$
Maximum value of the function is $f\left(1\right)=2$