Find the zeroes of the derivative?

  • Thread starter Thread starter negatifzeo
  • Start date Start date
  • Tags Tags
    Derivative
negatifzeo
Messages
66
Reaction score
0

Homework Statement


(3x^3-2x^2)-4x on the interval [-1,1]
Find the value of x that yields the global minimum.


Homework Equations


The derivative is 6x+4x-3


The Attempt at a Solution


Im not really sure. Don't I just find the zeroes of the derivative? It asks me for the EXACT answer which means the quadratic formula solution, but I am not sure how to.. Get there.
 
Physics news on Phys.org


You can't always be sure that the minimum on [-1, 1] lies on a zero of the derivative. Find where the derivative is 0, plug those x values back into the equation, then test both ends.
 


The max and min of a function on a give set can occur at three kinds of points:
1) At a point in the interior of the set where the derivative does not exist
2) At a point in the interior of the set where the derivative is 0
3) On the boundary of the set- which in the case of an interval consists of the two endpoints.

In this problem the derivative always exist so there are no points of type (1).

The derivative of (3x^3-2x^2)-4x is NOT " 6x+4x-3", it is 9x2- 4x- 4. The points at which that is 0 are indeed irrational but you can still calculate them exactly, using the quadratic formula. Since the problem asks only for the value of x at which the function has a minimum value, that should be enough. Even with approximate values for the function value at those points, you might be able to show that one of them gives the minimum value, or that the minimum is at an endpoint.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top