Question

If $\text{f(x)}\,\text{=}\,{{\text{x}}^{3/2}}\,(3\text{x}-10),\,\text{x}\,\ge \,0$ then in which of the following intervals f(x) is decreasing?
A(−∞, 0) ∪ (0, ∞)
B(2, ∞)
C(−∞, −1) ∪ [1, ∞)
D(−∞, 0) ∪ [2, ∞)

Answer 1

Since, $\text{f(x)}\,\text{=}\,{{\text{x}}^{3/2}}\,(3\text{x}-10),\,\text{x}\,\ge \,0$ we will find f1(x). We will equate f1(x) = 0 to find points where f(x) changes sign. Then, draw a wavy graph to see where the function is decreasing.

Complete step-by-step answer:
We will first try to understand the concept of decreasing function.
So, a function is decreasing on an interval if $\text{f(}{{\text{x}}_{1}}\text{)}\,\ge \,\text{f(}{{\text{x}}_{2}})$. Thus, a decreasing interval may also contain points where the function has a constant value.
Now, we have
$\text{f(x)}\,\text{=}\,{{\text{x}}^{3/2}}\,(3\text{x}-10),\,\text{x}\,\ge \,0$
Now, we will find the value of ${{\text{f}}^{\text{1}}}\text{(x)}$
$\begin{aligned} & {{\text{f}}^{1}}(\text{x})=\dfrac{\text{d}}{\text{dx}}\left( \text{f}(\text{x}) \right)=\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{3/2}}(3\text{x}-10) \right) \\\ & =\dfrac{\text{d}}{\text{dx}}\left( \text{3}{{\text{x}}^{{5}/{2}\;}}-\text{10}{{\text{x}}^{{3}/{2}\;}} \right) \\\ & =\dfrac{15}{2}{{\text{x}}^{{3}/{2}\;}}-15{{\text{x}}^{\dfrac{1}{2}}} \\\ & =\dfrac{15}{2}\sqrt{\text{x}}\,(\text{x}-2) \end{aligned}$
For a f(n) to be decreasing:-
${{\text{f}}^{1}}\text{(x)}\,\le \,0$
⇒ $\dfrac{15}{2}\sqrt{\text{x}}(\text{x}-2)\le 0$
⇒ $\sqrt{\text{x}}(\text{x}-2)\le 0$
But $\sqrt{\text{x}}\ge 0$ always
$\begin{aligned} & \Rightarrow \text{x}-2\le 0 \\\ & \Rightarrow \,\text{x}\le 2\,\text{but}\,\text{x}\,\ge \,0 \\\ \end{aligned}$
None of the options matches with the soln.

Note: We have used the formula.
$\dfrac{{\text{d}}}{{{\text{dx}}}}({\text{a}}{{\text{x}}^{\text{n}}}) = {\text{an}}{{\text{x}}^{{\text{n}} - 1}}$ to calculate $f^1(x)$.
Also, if a function f(x) is increasing, we must have $f^1(x)$ ≥ 0 & if a function is decreasing, we must have f1(x) ≤ 0.