5

What is an effective way to find that if the function, $f(x)=\sqrt{9x^2 +173x+900}- \sqrt{9x^2 +77x+900}$ is increasing or decreasing for $x \in R^+$

I was asked to find the range of $f(x)$ for $x \in \mathbb R^+$. I could find the limit of f(x) to be 16 at positive infinity. But for me to suggest that it would be its maximum and $f(0^+)$ would be its minimum, would require me to say that the function is increasing.

I was able to convert f(x) into: $\frac{96 }{\sqrt{9+ \frac{173}{x}+\frac{900}{x^2}}+\sqrt{9+\frac{77}{x}+\frac{900}{x^2}}}$.

But I'm still not able to "see" if it is increasing.

The derivative of $f(x)$ looks like a hassle, so I haven't really thought of it.

Thomas Andrews
  • 177,126
q123LsaB
  • 327
  • 1
  • 12
  • The derivative is quite easy (chain rule) and then you could calculate $f'(x)=0$ and see that there is only one solution at $-11.96$. – garondal Aug 11 '22 at 12:28
  • since square roots are in the equation i was reluctant to try to find the derivative. It would be nice if someone shows it. – q123LsaB Aug 11 '22 at 12:34

2 Answers2

3

Using your function form $\frac{96 }{\sqrt{9+ \frac{173}{x}+\frac{900}{x^2}}+\sqrt{9+\frac{77}{x}+\frac{900}{x^2}}}$, when $x>0$ and $x$ increases, both terms in the denominator decrease, therefore $f(x)$ increase.

ahala
  • 3,020
0

Because $g(x)=\sqrt{x}$ is a strictly increasing function, if $b>a$ then $\sqrt{b}>\sqrt{a}$. This implies that $\sqrt{96x+9x^2+77x+900}$ will grow faster than $\sqrt{9x^2+77x+900}$ for $x > 0$

Vasili
  • 10,690