I want to find whether the expression $D = \sqrt{5t^2 - 40t+125}$ is increasing or decreasing when $t=5$.

I want to find whether the expression $D = \sqrt{5t^2 - 40t+125}$ is increasing or decreasing when $t=5$.

My logic is I want to find whether is $f'(5)>0$ or $f'(5) < 0$.

I need to use the chain rule $h'(x) = g'(f(x))f'(x)$

$g'(f(x)) = \frac{1}{2}(5t^2 - 40t+125)^\frac{-1}{2}$

$f'(x) = 10t-40$

$h'(x) = \frac{1}{2}(5t^2 - 40t+125)(10t-40)^\frac{-1}{2}$

This is a non calculator paper and is this really possible without a calculator?

Your expression has a slight error. It should be:

$h'(t) = \frac{1}{2}(5t^2 - 40t+125)^\frac{-1}{2}(10t-40)$

Since all you need to know is the sign of $h'(t)$, you only need to look at the sign of $5t^2 - 40t+125$ for $t=5$ and $10t-40$ when $t=5$.

Since the last term is clearly $>0$, the sign of $5t^2 - 40t+125$ at $t=5$ tells you if $h'$ is greater than or less than 0

• what do you mean by, all you need to know is the sign of $h'(t)$ Commented May 2, 2016 at 5:40
• @dagda the function is increasing when the derivative (which you call h') is greater than $0$ and decreasing when it is less than 0. Commented May 2, 2016 at 5:42
• and I can tell this from only looking at the last term? Commented May 2, 2016 at 5:43
• @dagda1 Well the last term is positive at $t=5$ and the first term is 0.5 which is positive. So a positive * positive number is positive, and a negative * positive number is negative, so that middle term tells you if the entire product is positive or negative Commented May 2, 2016 at 5:44

$D = \sqrt{5t^2 - 40t+125}=\sqrt{5(t^2-8t+25)}$

Let $f(t)=t^2-8t+25=(t-4)^2+9$

If $t\ge4$ and $t_1<t_2$ then $f(t_1)<f(t_2)$

• I don't get how you simplified the square root or how there is no longer a square root and the 5 has disappeared Commented May 2, 2016 at 5:36
• $\sqrt x$ is strictly increasing Commented May 2, 2016 at 5:39

Because $\sqrt{x}$ is strictly increasing, this question is equivalent to whether $f(t)=D^2=5t^2-40t+125$ is increasing or decreasing at $t=5$, $f'(5)=10>0$ so it's increasing at $t=5$.