How do you find all local maximum and minimum points using the second derivative test given $y = x^{3} - 9 x^{2} + 24 x$ $y=x^3-9x^2+24x$ ?

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Answer 1 · 2017-03-16T14:39:45

We have a local maximum at $(2, 20)$ $(2,20)$
We have a local minimum at $(4, 16)$ $(4,16)$

Let's calculate the first derivative

$y = x^{3} - 9 x^{2} + 24 x$ $y=x^3-9x^2+24x$

Let $f (x) = x^{3} - 9 x^{2} + 24 x$ $f(x)=x^3-9x^2+24x$

$f'(x)=dy/dx=3x^2-18x+24$

$=3(x^2-6x+8)=3(x-2)(x-4)$

The critical points are when $dy/dx=0$

That is, when $x=2$ and $x=4$

Now, we calculate the second derivative

f''(x)=(d^2y)/dx^2=6x-18#

When $f''(x)=(d^2y)/dx^2=0$ , we have an inflexion point at $x=3$

Now,

We calculate

$f''(2)=12-18=-6$

As, $f''(2) <0$ , we have a local maximum at $(2,20)$

$f''(4)=24-18=6$

As, $f''(2) >0$ , we have a local minimum at $(4,16)$

graph{x^3-9x^2+24x [-36.67, 45.55, -6.95, 34.14]}

How do you find all local maximum and minimum points using the second derivative test given y=x^3-9x^2+24xy=x3−9x2+24xy=x^3-9x^2+24x?