# Let f(x)=(5/2)sqrt(x). The rate of change of f at x=c is twice its rate of change at x=3. What is the value of c?

Nov 2, 2016

We start by differentiating, using the product rule and the chain rule.

Let $y = {u}^{\frac{1}{2}}$ and $u = x$.

$y ' = \frac{1}{2 {u}^{\frac{1}{2}}}$ and $u ' = 1$

$y ' = \frac{1}{2 {\left(x\right)}^{\frac{1}{2}}}$

Now, by the product rule;

$f ' \left(x\right) = 0 \times \sqrt{x} + \frac{1}{2 {\left(x\right)}^{\frac{1}{2}}} \times \frac{5}{2}$

$f ' \left(x\right) = \frac{5}{4 \sqrt{x}}$

The rate of change at any given point on the function is given by evaluating $x = a$ into the derivative. The question says that the rate of change at $x = 3$ is twice the rate of change at $x = c$. Our first order of business is to find the rate of change at $x = 3$.

$r . c = \frac{5}{4 \sqrt{3}}$

The rate of change at $x = c$ is then $\frac{10}{4 \sqrt{3}} = \frac{5}{2 \sqrt{3}}$.

$\frac{5}{2 \sqrt{3}} = \frac{5}{4 \sqrt{x}}$

$20 \sqrt{x} = 10 \sqrt{3}$

$20 \sqrt{x} - 10 \sqrt{3} = 0$

$10 \left(2 \sqrt{x} - \sqrt{3}\right) = 0$

$2 \sqrt{x} - \sqrt{3} = 0$

$2 \sqrt{x} = \sqrt{3}$

$4 x = 3$

$x = \frac{3}{4}$

So, the value of $c$ is $\frac{3}{4}$.

Hopefully this helps!