Question

How do you find the max or minimum of `f(x)=x^2-8x+2`?

Answer 1

Check the coefficient of `x^2`, then calculate the `y`-coordinate of the vertex to get `min = -14`.

Explanation:

A quadratic equation in standard form is `y = ax^2 + bx + c`. Here, `a = 1 > 0`, which means that the parabola must be facing upwards, and that it has a minimum instead of a maximum bound.

To calculate the `y`-coordinate of the vertex, one can find the `x`-coordinate first then substitutue it into the equation, among other ways, but I would personally use the following:

`k = c - (b^2)/(4a)`

We know that `a = 1`, `b = -8` and `c = 2`, so let's substitute:

`k = 2 - ((-8)^2)/(4(1)) = 2 - 64/4 = 2 - 16 = -14`.

So, we have `min = -14`. Let's check:

graph{(x^2 - 8x + 2 - y)(-14 -y)=0 [-11.82, 20.22, -15.38, 0.64]}

Answer 2

The first derivative zero is the extrema. The second derivative sign indicates the direction, or use the sign of the first derivative on each side.

Explanation:

`f(x) = x^2 - 8x + 2`

`f'(x) = 2x - 8` Zero at `x = 4`
Negative at `x = 3`, Positive at `x = 5` = Minimum

First Derivative Test
Suppose f(x) is continuous at a stationary point x_0.

1. If `f^'(x)>0` on an open interval extending left from x_0 and `f^'(x)<0` on an open interval extending right from x_0, then f(x) has a local maximum (possibly a global maximum) at x_0.

2. If `f^'(x)<0` on an open interval extending left from x_0 and `f^'(x)>0` on an open interval extending right from x_0, then f(x) has a local minimum (possibly a global minimum) at x_0.

3. `If f^'(x)` has the same sign on an open interval extending left from x_0 and on an open interval extending right from x_0, then f(x) has an inflection point at x_0.

Weisstein, Eric W. "First Derivative Test." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/FirstDerivativeTest.html
REFERENCES:
Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 14, 1972.
Referenced on Wolfram|Alpha: First Derivative Test

Second Derivative Test

Suppose f(x) is a function of x that is twice differentiable at a stationary point x_0.

1. If f^('')(x_0)>0, then f has a local minimum at x_0.

2. If f^('')(x_0)<0, then f has a local maximum at x_0.

The extremum test gives slightly more general conditions under which a function with f^('')(x_0)=0 is a maximum or minimum.

If f(x,y) is a two-dimensional function that has a local extremum at a point (x_0,y_0) and has continuous partial derivatives at this point, then f_x(x_0,y_0)=0 and f_y(x_0,y_0)=0. The second partial derivatives test classifies the point as a local maximum or local minimum.

Define the second derivative test discriminant as
D = f_(xx)f_(yy)-f_(xy)f_(yx)
= f_(xx)f_(yy)-f_(xy)^2.

Then

1. If D>0 and f_(xx)(x_0,y_0)>0, the point is a local minimum.

2. If D>0 and f_(xx)(x_0,y_0)<0, the point is a local maximum.

3. If D<0, the point is a saddle point.

4. If D=0, higher order tests must be used.

REFERENCES:
Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 14, 1972.