1. ## Increasing Function

Find all exact values of $a$ for which $f(x) = (x^2 + ax -1)\sqrt{2x-3}$ is always an increasing function.

My working:

First notice the domain: $x \in [ \frac{3}{2} , \infty)$

We require all $f'(x) > 0$ for all x within this domain.

Thus $f'(x) = \frac{5x^2 + 3(a-2)x - 3a - 1}{\sqrt{2x-3}}$

Here we place a restriction on the original domain for $x \in (\frac{3}{2} , \infty)$ [since $x \neq \frac{3}{2}$, because derivative is not defined at this point]

Thus $\sqrt{2x-3} > 0$ for all $x \in (\frac{3}{2} , \infty)$

Thus we now need to ensure $5x^2 + 3(a-2)x - 3a - 1 > 0$

Solving for x = 0 yields:

$x = \frac{-3(a-2) \pm \sqrt{9a^2+24a+56}}{10}$

Notice $\Delta > 0$ ie $9a^2+24a+56 > 0$ for all a

Now what?

2. Originally Posted by usagi_killer
Find all exact values of $a$ for which $f(x) = (x^2 + ax -1)\sqrt{2x-3}$ is always an increasing function.

My working:

First notice the domain: $x \in [ \frac{3}{2} , \infty)$

We require all $f'(x) > 0$ for all x within this domain.

Thus $f'(x) = \frac{5x^2 + 3(a-2)x - 3a - 1}{\sqrt{2x-3}}$

Here we place a restriction on the original domain for $x \in (\frac{3}{2} , \infty)$ [since $x \neq \frac{3}{2}$, because derivative is not defined at this point]

Thus $\sqrt{2x-3} > 0$ for all $x \in (\frac{3}{2} , \infty)$

Thus we now need to ensure $5x^2 + 3(a-2)x - 3a - 1 > 0$

Solving for x = 0 yields:

$x = \frac{-3(a-2) \pm \sqrt{9a^2+24a+56}}{10}$

Notice $\Delta > 0$ ie $9a^2+24a+56 > 0$ for all a

Now what?
Those two points divide the number line into three intervals in each of which the derivative has the same sign. Calculate the derivative at one point in each interval to determine whether that sign is "+" or "-".