## IB Maths SL Second Derivative test

Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths SL Revision Notes

### IB Maths SL Second Derivative test

IB Mathematics SL – Derivatives, Turning points, 2nd derivative test

How can we find coordinates and nature of all the stationary points of the function $f(x)=x^3-12x+2$ ?

Thanks
Oliver

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Joined: Mon Jan 28, 2013 8:08 pm

### Re: IB Maths SL Second Derivative test

IB Mathematics SL – Derivatives, monotonicity, increasing functions, turning points

We know that

If $f'(x_{0})=0$ and $f''(x_{0})>0$ then at $x= x_{0}$ the function has a minimum turning point

If $f'(x_{0})=0$ and $f''(x_{0})<0$ then at $x= x_{0}$ the function has a maximum turning point

If $f'(x_{0})=0$ and $f''(x_{0})=0$ and the second derivative function changes sign around $x_{0}$ then at $x= x_{0}$ the function has a stationary point of inflection.

The first derivative function is

$f'(x)=3x^2-12$

By setting the derivative equal to zero,

$f'(x)= 3x^2=12\Rightarrow x=\pm 2$

So we know that there are stationary points when $x=2 \or\ x=-2$.

By substituting $x$ into the original equation

$f(2)=2^3-12(2)+2=-14$

and

$f(-2)=(-2)^3-12(-2)+2=18$

Thus the coordinates of the stationary points are $(2,-14) \and\ (-2,18)$

The second derivative function is

$f''(x)=6x$

By substituting the $x$ values of the stationary points into second derivative function

$f''(2)=6(2)=12>0$, so is a local minimum point

and

$f''(-2)=6(-2)=-12<0$ , so is a local maximum point

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: IB Maths SL Second Derivative test

thank you miranda!
Oliver

Posts: 0
Joined: Mon Jan 28, 2013 8:08 pm