Problem working out a second derivative I am studying maths as a hobby and feel I am having problems working out second derivatives.
The problem is as follows:
Find the maximum or minimum values of the function $y = (2x - 5)^4$
Here is my working:
$$dy/dx = 4(2x - 5)^3\cdot 2 = 8(2x -5)^3 ,$$
Which is zero when $x = 2\frac {1}{2}$.
To find whether this is a maximum or a minimum I find the second derivative:
$$d^2y/dx^2 = 24(2x - 5)^2.2 = 48(2x - 5)^2.$$
But this is where I feel I must have gone wrong because this is zero when $x = 2\frac {1}{2}$.
 A: The second derivative test, that is, computing the second derivative of $f$ does not always tell you whether or not a point is a local maximum or local minimum. It is only always true that if $f''(a) > 0$, then $f$ has a local minimum at $a$.
Hence, when you encounter (as in this case) $f''(a) = 0$, a simple alternative would be to fall back to inspecting the first derivative in an interval around the critical point. For instance, if $x < a \Rightarrow f'(x) < 0$ and $x > a \Rightarrow f'(x) > 0$, then we can safely conclude that the critical point at $x=a$ is a local minimum.
The computations should be trivial given that you have worked out the value of $x = 2\frac{1}{2}$ to be a stationary point.
A: You are correct, so in this case second derivative doesn't tell us anything about whether the point is a maximum, minimum or an inflection point.
But what does the function $y=(2x-5)^4$ look like? Sketch it.
Polynomials like $x^4$ or $-x^4$ have derivatives equal to zero at the critical points, but it is a minimum for $x^4$ and maximum for $-x^4$.
A: If you insist on the criterion for extrema using higher derivatives -- which may not be wise all the time -- you find first the zeros of the first derivative.  Now substitute that zero in the higher derivatives of $f$ until it's not a zero of that derivative.  If the order of  this derivative is even, there'e an extremum of $f$, if the order is even we have an inflection point.
A: If the second derivative test fails, I usually check the first derivative at points enclosing the point of interest, e.g. 2 and 3
since the first derivative at x=2 is negative and that at x=3 is positive, I will conclude that it is a minima at x=2.5
