Question

Find the value of $x,$ if $f\left( x \right)=2{{x}^{3}}-15{{x}^{2}}-84x-7$ is a decreasing function.

Answer 1

We will use the first derivative test for decreasing functions to find the value of the unknown variable $x.$ By first derivative test, the function $f$ is decreasing if ${f}'\left( x \right)<0.$ We will find the first derivative of the given function and then we will find the values of $x$ for which the first derivative of the given function is less than $0.$

Complete step by step solution:
Let us consider the given function $f\left( x \right)=2{{x}^{3}}-15{{x}^{2}}-84x-7$
We are asked to find the values of $x$ for which the given function is a decreasing function.
We will use the first derivative test for decreasing functions to find the values of $x.$
According to the first derivative test for decreasing functions, a function $f$ is a decreasing function if its first derivative is less than zero, i.e., if ${f}'<0.$
Now, let us find the first derivative of the given function.
We will get ${f}'\left( x \right)=6{{x}^{2}}-30x-84$
As we can see, $6$ is the common factor and let us take it out. Then, we will get ${f}'\left( x \right)=6\left( {{x}^{2}}-5x-14 \right)$
Let us factorize the above polynomial by splitting the middle term.
We will get ${f}'\left( x \right)=6\left( {{x}^{2}}-5x-14 \right)=6\left( {{x}^{2}}-7x+2x-14 \right)$
Now, we will take $x$ out from the first two terms and $2$ from the last two terms.
We will get {f}'\left( x \right)=6\left\\{ x\left( x-7 \right)+2\left( x-7 \right) \right\\}
Now, we will take $x-7$ out. Then we will get ${f}'\left( x \right)=6\left( x-7 \right)\left( x+2 \right).$
From the first derivative test, we will get ${f}'\left( x \right)<0$
So, we will get $6\left( x-7 \right)\left( x+2 \right)<0$
After transposing $6,$ we will get $\left( x-7 \right)\left( x+2 \right)<0$
And we know that this is only possible if either $x-7<0$ and $x+2>0$ or $x-7>0$ and $x+2<0.$
Let us consider $x-7>0$ and $x+2<0$
Then, we will get $x>7$ and $x<-2.$
We know that this is not possible.
Now, we will consider $x-7<0$ and $x+2>0$
We will get $x<7$ and $x>-2$
We know that this is possible.

**Hence the given function is a decreasing function when $-2 < x < 7.$ **

Note:
Let us recall the first derivative test for increasing and decreasing functions. Suppose that $f$ is a function defined in an interval $I.$ If ${f}'<0,$ then the function is a decreasing function. If ${f}'>0,$ then the function is an increasing function. And if ${f}'=0$ or ${f}'$ does not exist, then the function has critical points at the points where ${f}'=0$ or ${f}'$ does not exist.