# Find an equation for the horizontal tangent to the curve y=x-3root x

• Oct 17th 2010, 05:08 AM
r-soy
Find an equation for the horizontal tangent to the curve y=x-3root x
Hi
a ) find an equation for the horizontal tangent to the curve y=x-3root x
b) What is range of values of values of curve's slope ?
c ) What is range of values of curve ?
number a already I solved but my queation now in b and c
-----
I try to solve
b)
to get curve's slope must derivatives
range of values of values of curve's slope is ( - infinity , 1 ]
becuse when we put ( 0 ) in curve's slope It will comw 1

c ) did here put zero in y=x-3root x

• Oct 17th 2010, 05:50 AM
skeeter
Quote:

Originally Posted by r-soy
Hi
a ) find an equation for the horizontal tangent to the curve y=x-3root x
b) What is range of values of values of curve's slope ?
c ) What is range of values of curve ?
number a already I solved but my queation now in b and c
-----
I try to solve
b)
to get curve's slope must derivatives
range of values of values of curve's slope is ( - infinity , 1 ]
becuse when we put ( 0 ) in curve's slope It will comw 1

c ) did here put zero in y=x-3root x

b) $\displaystyle y' = 1 - \frac{3}{2\sqrt{x}}$ ... range of the derivative is $\displaystyle -\infty < y' < 1$ ... note that $\displaystyle y' \ne 1$
c) $\displaystyle y' = 0$ at $\displaystyle x = \frac{9}{4}$
$\displaystyle y'' = \frac{3}{4x\sqrt{x}} > 0$ for all $\displaystyle x$ in the domain of $\displaystyle y$, therefore $\displaystyle y$ has an absolute minimum at $\displaystyle x = \frac{9}{4}$ ...
$\displaystyle y\left(\frac{9}{4}\right) = -\frac{9}{4}$
so, range of $\displaystyle y$ is $\displaystyle y \ge -\frac{9}{4}$