Thread: Differentiation.. the values of x for which dy/dx > 0

Given that $\displaystyle y=x^2-6x+9 $, the values of x for which $\displaystyle \frac {dy}{dx} \geq 0 $

This is what I'm thinking.

First differentiate. Then put the sign then 0. You will get $\displaystyle 2x-6 \geq 0 $

Is this right?

Originally Posted by jgv115

Given that $\displaystyle y=x^2-6x+9 $, the values of x for which $\displaystyle \frac {dy}{dx} \geq 0 $

This is what I'm thinking.

First differentiate. Then put the sign then 0. You will get $\displaystyle 2x-6 \geq 0 $

Is this right?

Yes,
from there you find the range of x for which $\displaystyle f'(x)\ \ge\ 0$

In the case of quadratics, you can alternatively find the axis of symmetry.
The co-efficient of $\displaystyle x^2$ is positive, so the graph is U-shaped.
Hence the slope of the tangent is $\displaystyle \ge\ 0$ at or after the minimum,
which occurs halfway between the roots or at the roots in the case of a double root.

$\displaystyle x^2-6x+9=(x-3)(x-3)$

Double root at x=3, so the tangent has a positive slope after x=3.