Find a local minimum of the function f(x) = x^3 - 2x.

To find a local minimum (i.e. a point where the function changes from a negative slope into a positive slope), we first need to find all points where the slope of the function is zero. The first derivative of a function gives information about the slope, so we find the first derivative ( f'(x) = 3x2 - 2 ) and set it equal to zero. 0 = 3x2 - 2 can be rearranged to x2 = 2/3. Taking the square root of both sides of the equation the slope is zero in the two points at approx. x = 0.816 and x = -0.816. In order to find out which of the two extrema is a minimum (not a maximum), we could simply check the slope just before and just after our two candidate points and confirm if the slope is indeed negative just before the extremum and positive just after. Alternatively we can find the rate of change of the slope in the two candidate points using the second derivative of f (f''(x) = 6x). For x = 0.816 the second derivative is f''(0.816) = 4.899. This positive rate of change indicates that the slope is increasing at this point. In other words, the slope at this moment is zero, but it is currently changing in a positive direction, so changing from a lower (a negative) value to a higher (a positive value). Finally we simply find the y-value by plugging x back into the original function --> f(0.816) = -1.09. We can conclude that the minimum can be found at (0.816 | -1.09).

KH
Answered by Karoline H. Maths tutor

3673 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The height (h) of water flowing out of a tank decreases at a rate proportional to the square root of the height of water still in the tank. If h=9 at t=0 and h=4 at t=5, what is the water’s height at t=15? What is the physical interpretation of this?


The line L1 has vector equation,  L1 = (  6, 1 ,-1  ) + λ ( 2, 1, 0). The line L2 passes through the points (2, 3, −1) and (4, −1, 1). i) find vector equation of L2 ii)show L2 and L1 are perpendicular.


A curve has the equation y = 4x^3 . Differentiate with respect to y.


Integrate (x^3 - x^2 - 5x + 7) with respect to x.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning