# Thread: Show that the function...

1. ## Show that the function...

Show that the function y=x³(1-x)⁵ has a horizontal tangent at point P with a x coordinate 3/8. Show that the y coordinate of P is 3³ x 5⁵ /8⁸.

Sorry guys I have absolutely no idea. Was thinking along the lines of the product rule, but not sure. Help with working out appreciated.

Cheers

2. The product rule is right. What are you getting when you try to apply the product rule?

3. y=x³(1-x)⁵

y¹=3x²(1-x)⁵ +15x⁴(1-x)⁴ ???

4. Originally Posted by Joel
y=x³(1-x)⁵

y¹=3x²(1-x)⁵ +15x⁴(1-x)⁴ ???
$\displaystyle y=x^3(1-x)^5$

Let $\displaystyle f(x)=x^3$

Then $\displaystyle f'(x)=3x^2$

Let $\displaystyle g(x)=(1-x)^5$

Then $\displaystyle g'(x)=5(1-x)^4(-1)$

$\displaystyle y=f(x)g(x)$

$\displaystyle y\,'=f(x)g'(x)+g(x)f'(x)$

So what would you get for $\displaystyle y\,'$?

To show that $\displaystyle y$ has a horizontal tangent line at $\displaystyle x=\frac{3}{8}$, you need to show that $\displaystyle y\,'=0$ when $\displaystyle x=\frac{3}{8}$