Question

Find the local max and min for \[{x^3} - 27x\] .

Answer 1

Hint: The equation should be rewritten as a function of x first. Then the first and second derivative should be obtained of the rewritten function. The derivative should be set equal to zero for further solution. Then further simplification should be done. The complete solution is the result of both the positive and negative portions of the solution.

Complete step-by-step answer:
Rewriting the given function \[{x^3} - 27x\] as a function of “x” we have,
 \[f\left( x \right) = {x^3} - 27x\]
Now, finding the first derivative of the function by differentiating it using the power rule which states that,
 \[\dfrac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}}\] Where “n=3”
Hence, we have
 \[
   \Rightarrow 3{x^2} + \dfrac{d}{{dx}}\left[ { - 27x} \right] \\
   \Rightarrow 3{x^2} - 27 \;
 \]
Now, finding the second derivative of the function by the sum rule, the derivative of \[3{x^2} - 27\] with respect to “x” is \[\dfrac{d}{{dx}}\left[ {3{x^2}} \right] + \dfrac{d}{{dx}}\left[ { - 27} \right] = 6x\]
To find the local maximum and minimum values of the function, the derivatives have to be set to zero and then solved. Hence, we have
 \[3{x^2} - 27 = 0\]
Adding 27 to both sides of the equation we have,
 \[ \Rightarrow 3{x^2} = 27\]
Dividing each term by 3 and simplifying we have,
 \[
  {x^2} = \dfrac{{27}}{3} \\
   \Rightarrow {x^2} = 9 \;
\]
Taking square root of both sides of the equation to eliminate the exponent on the left side
 \[ \Rightarrow x = \pm \sqrt 9 \]
The complete solution is the result of both the positive and negative portions of the solution
 \[
   \Rightarrow x = \pm 3 \\
   \Rightarrow x = 3, - 3 \;
 \]
Evaluating the second derivative at \[x = 3\] ,
If the second derivative is positive, then this is a local minimum. If it is negative then this is a local maximum. Multiplying 6 and 3 that is 18.
The local minimum is \[x = 3\] because the value of the second derivative is positive. This is referred to as the second derivative test.
Now finding the “y” value when \[x = 3\] then
 \[ \Rightarrow y = - 54\]
Evaluating the second derivative at \[x = 3\] ,
If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum. Multiplying 6 and -3 we have -18.
Local maximum is \[x = - 3\] because the value of the second derivative is negative. This is referred to as the second derivative test.
The value of y is found out when \[x = - 3\]
 \[
   \Rightarrow f\left( { - 3} \right) = {\left( { - 3} \right)^3} - 27 \times \left( { - 3} \right) \\
   \Rightarrow f\left( { - 3} \right) = 54 \\
   \Rightarrow y = 54 \;
 \]
These are the local extrema for \[{x^3} - 27x\] .
 \[\left( {3, - 54} \right)\] Is a local minima
 \[\left( { - 3,54} \right)\] Is a local maxima
So, the correct answer is “ \[\left( {3, - 54} \right)\] Is a local minima and
 \[\left( { - 3,54} \right)\] Is a local maxima”.

Note: Finding the first derivative and second derivative of the function by differentiating it using the power rule is important. To find the local maximum and minimum values of the function, the derivatives have to be set to zero and then solved. If the second derivative is positive, then this is a local minimum. If it is negative then this is a local maximum.