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

1. ## 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

please help me .

2. 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

please help me .
b) $y' = 1 - \frac{3}{2\sqrt{x}}$ ... range of the derivative is $-\infty < y' < 1$ ... note that $y' \ne 1$

c) $y' = 0$ at $x = \frac{9}{4}$

$y'' = \frac{3}{4x\sqrt{x}} > 0$ for all $x$ in the domain of $y$, therefore $y$ has an absolute minimum at $x = \frac{9}{4}$ ...
$y\left(\frac{9}{4}\right) = -\frac{9}{4}$

so, range of $y$ is $y \ge -\frac{9}{4}$