Question

Find the intervals in which the function \[f(x) = \dfrac{{{x^4}}}{2} - {x^3} - 5{x^2} + 24x + 12\] is:
A. strictly increasing
B. strictly decreasing

Answer 1

Hint: According to the given question, firstly differentiate the function with respect to x and then put \[f'(x) = 0\] and simplify to calculate the initial points.Then check if \[f'(x) > 0\] then the function is strictly increasing and if \[f'(x) < 0\] then the function is strictly decreasing.

Complete step-by-step solution:
 The function is given as:
\[f(x) = \dfrac{{{x^4}}}{2} - {x^3} - 5{x^2} + 24x + 12\]
First, we need to differentiate the above function to obtain the first derivative, denoted by \[{f'}(x)\]
\[f'(x) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^4}}}{4} - {x^3} - 5{x^2} + 24x + 12} \right)\]
  (∵ The derivative of a constant is zero.)
\[ \Rightarrow \dfrac{1}{4}\dfrac{d}{{dx}}{x^4} - \dfrac{d}{{dx}}{x^3} - 5\dfrac{d}{{dx}}{x^2} + 24\dfrac{{dx}}{{dx}} + 0\]
On simplifying,
\[ \Rightarrow \dfrac{1}{4}.4{x^3} - 3{x^2} - 5.2x + 24.1 + 0\]
So, we get,
\[f'(x) = {x^3} - 3{x^2} - 10x + 24\]
Now, to take the factor for the function, we take x to be numbers starting from 1 until \[{f'}(x)\] comes out to be equal to 0.
By trial method, we find that \[x = 2\] gives \[{f'}(x) = 0\]. Thus, the factor is \[(x - 2)\]
We then factorize \[f'(x)\] by \[(x - 2)\] using a long form division method.
\[\begin{array}{l} \Rightarrow \dfrac{{{x^3} - 3{x^2} - 10x + 24}}{{x - 2}}\\ \Rightarrow {x^2} - x - 12\end{array}\]
\[ \Rightarrow (x + 3)(x - 4)\]
Thus, \[f'(x) = (x - 2)(x + 3)(x - 4)\]
Hence, the initial points here are 2, -3 and 4. Now, we place these points on a real number line with limits from \[ - \infty \] to \[ + \infty \] .

Thus, the intervals are \[\left( { - \infty , - 3} \right),\left( { - 3,2} \right),\left( {2,4} \right)\] and \[\left( {4,\infty } \right)\]
A. For \[f(x)\] to be strictly increasing, we must have
\[f'(x) > 0\]
\[\begin{array}{l} \Rightarrow (x - 2)(x + 3)(x - 4) > 0\\ \Rightarrow x \in ( - 3,2) \cup (4,\infty )\end{array}\]
Since the value of \[f'(x) > 0\] or strictly increasing in both the intervals.
We can also get this idea from the number line where the intervals \[\left( { - 3,{\rm{ }}2} \right)\] and \[\left( {4,{\rm{ }}\infty } \right)\] will be strictly increasing.

B. For \[f(x)\] to be strictly decreasing, we must have
\[f'(x) < 0\]
\[\begin{array}{l} \Rightarrow (x - 2)(x + 3)(x - 4) < 0\\ \Rightarrow x \in ( - \infty , - 3) \cup (2,4)\end{array}\]
So, \[f(x)\] is strictly decreasing on \[x \in ( - \infty , - 3) \cup (2,4)\]
As shown in the number line the intervals \[\left( { - \infty , - 3} \right)\] and \[\left( {2,4} \right)\] will be strictly decreasing.

Note: To solve these types of questions, your first step is to find the derivative of the given function and then apply all the conditions for which we have to verify. You can also draw the number line for better representation of the strictly increasing and decreasing function represented as \[ + \] and \[ - \] respectively as shown above.