Question

I’m having trouble doing this calculus problem from my prep guide Need help

Answer 1

The given inequality is:

$-18\sqrt[]{2x-1}>-36$

To find the solution start by dividing both sides by 18:

$\begin{gathered} \frac{-18\sqrt[]{2x-1}}{18}>-\frac{36}{18} \\ -\sqrt[]{2x-1}>-2 \end{gathered}$

Now, multiply both sides by -1:

$\begin{gathered} -\sqrt[]{2x-1}\cdot-1>-2\cdot-1 \\ As\text{ we are multiplying by a negative number, reverse the inequality symbol:} \\ \sqrt[]{2x-1}<2 \end{gathered}$

Square both sides:

$\begin{gathered} \sqrt[]{2x-1}^2<2^2 \\ 2x-1<4 \end{gathered}$

Add 1 to both sides:

$\begin{gathered} 2x-1+1<4+1 \\ 2x<5 \end{gathered}$

Divide both sides by 2:

$\begin{gathered} \frac{2x}{2}<\frac{5}{2} \\ x<\frac{5}{2} \\ x<2.5 \end{gathered}$

Also, the number inside the square root can't be negative, then:

$\begin{gathered} 2x-1\ge0 \\ 2x\ge1 \\ x\ge\frac{1}{2} \\ x\ge0.5 \end{gathered}$

Then 0.5<=x<2.5, in the number line it will be:

The circle in 0.5 is closed and the circle in 2.5 is open.

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