# Thread: Math help wanted :) (Finding stationary points etc.)

1. ## Math help wanted :) (Finding stationary points etc.)

Let
f(x) = 3x^3 - 18 squarerootx
for 0 equal/less x equal/less 4.

(a) Find any local maxima, minima or points of inflexion.
(b) Sketch the function, clearly marking and labeling the domain and range
and the global maximum and minimum.

-18 is not little, it's just a regular -18.

2. $y= 3x^3-18\sqrt{x}$

$\dfrac{dy}{dx}= 9x^2-\dfrac{9}{\sqrt{x}}$

For min/max values solve for

$\dfrac{dy}{dx}= 0$

$9x^2-\dfrac{9}{\sqrt{x}}=0$

3. So x is 1?
and then what do I do?

Sorry I wrote 9 before that was a typo

4. $x \neq 9$

Once you find the true value for x you then sub that back into the equation y.

Spoiler:
$x = 1$

5. Yup I know it equals 1 sorry it was a typo before. So subbing 1 back into the original equation I got =-15 so that means there is a local maximum there? My textbook says to find the 2nd derivative and then sub the x = 1 into that derivative. if y'' < 0 it is a maximum, etc etc
So what is the 2nd derivative? 18x... what is the rest? Square roots confuse me!

6. Originally Posted by brumby_3
Yup I know it equals 1 sorry it was a typo before. So subbing 1 back into the original equation I got =-15 so that means there is a local maximum there?
What makes you think it is a maximum?
My textbook says to find the 2nd derivative and then sub the x = 1 into that derivative. if y'' < 0 it is a maximum, etc etc
So what is the 2nd derivative? 18x... what is the rest? Square roots confuse me!
Then think $\sqrt{x}= x^{\frac{1}{2}}$.
That, I suspect, was how pickslides got the first derivative: $y= 3x^3- 18x^\frac{1}{2}$ so $y'= 3(3x^{3-1})- 18\left(\frac{1}{2}x^{\frac{1}{2}- 1}\right)= 9x^2- 9x^{-\frac{1}{2}}$.
(And, of course, $x^{-\frac{1}{2}}= \frac{1}{\sqrt{x}}$.)

Now, $y''= 9(2x^{2-1})- 9(-\frac{1}{2})x^{-\frac{1}{2}-1}$
What is that?