1. ## critical points

Hi;
Do all critical points ie local max/local min have a slope of zero.

Thanks.

2. ## Re: critical points

No. This is wrong for two reasons. First a "critical point" is defined in any Calculus text as a point at which the derivative is 0 or does not exist. For example, the absolute value function, y= |x|, has derivative equal to 1 for x> 0, -1 for x< 0, never 0. x= 0 is, however, a critical point because the derivative does not exist there. Second your "ie local max/local min" is wrong because a critical point is not necessarily either a local maximum or a local minimum. $\displaystyle y= x^3$ has derivative $\displaystyle y'= 3x^2$ which is 0 at x= 0 so x= 0 is a critical point. But $\displaystyle y= x^3$ has neither a local maximum not a local minimum there.

3. ## Re: critical points

Ok so why isnt my tangent zero (tis is my first look at derivatives and calculus) so need some help.

I posted my graph to help.

Thanks.

4. ## Re: critical points

Looks like $f(x)=x^3-4x^2-5x+36$

So, $f'(x)=3x^2-8x-5$

$3x^2-8x-5=0$

Two critical values ...

$x=\dfrac{4 \pm 4\sqrt{2}}{3}$

At $x=\dfrac{4+4\sqrt{2}}{3} \approx 3.21895$, $y \approx 11.8123$

tangent line at that point is $y=11.8123$, as shown in the attached graph.

How did you arrive at your tangent line equation?

5. ## Re: critical points

Thanks skeeter but that equation x^3 - 4x^2 - 5x + 36 should be x^3 - 4x^2 - 9x + 36.

So derivative 3x^2 - 8x - 9.

6. ## Re: critical points

Ok ... so the criticsl values are $x=\dfrac{4 \pm \sqrt{43}}{3}$

at $x=\dfrac{4+\sqrt{43}}{3} \approx 3.51915$, $y \approx -1.6274$

Same drill ... tangent line is $y=-1.6274$. new graph attached

Next time post the function and what you were supposed to find (i.e., directions) ... the thumbnail you posted isn't that large even after clicking on it.

7. ## Re: critical points

setting the derivative = 0 I get x = 3.49?

8. ## Re: critical points

Originally Posted by anthonye
setting the derivative = 0 I get x = 3.49?
How did you arrive at that value?

9. ## Re: critical points

Ok

3x^2 - 8x - 9 = 0

3x^2 - 8x = 9

x^2 - 8/3x = 3

x^2 - 8/3x + 1.8 = 4.8

(x - 1.3)(x - 1.3) = 4.8

(x - 1.3)^2 = 4.8

(x - 1.3) = + or - 2.19

x = 2.19 + 1.3 = 3.49
x = 2.19 - 1.3 = -0.89.

10. ## Re: critical points

Note that $\left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9} \ne 1.8$. ... your calculation is off because you rounded too soon. Put the calculator away until the very end ...

$x^2 - \dfrac{8}{3}x = 3$

$x^2 - \dfrac{8}{3}x + \left(\dfrac{4}{3}\right)^2 = 3+\left(\dfrac{4}{3}\right)^2$

$\left(x-\dfrac{4}{3}\right)^2 = \dfrac{43}{9}$

$x-\dfrac{4}{3} = \pm \dfrac{\sqrt{43}}{3}$

$x = \dfrac{4 \pm \sqrt{43}}{3}$

NOW use the calculator ...

11. ## Re: critical points

Ok thanks skeeter its correct now, thanks for all your help.