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