Optimizing f(x)

Summary and examples

f'(a) is the derivative of a function f(x) in a point (a,f(a)) and is the slope of the tangent line at that point of the graph of that function.

When f'(a)=0 it means that the tangent line in (a, f(a)) has a slope 0 and thus that it is horizontal to the X-axis. For such a point there are three possibilities:

  1. the point is a (local) maximum;
  2. the point is a (local) minimum;
  3. the point is a horizontal inflection point (an example is the graph of y=x^3).

When we want to know in which points a function f(x) has a local maximum, minimum or horizontal inflection point, we calculate the derivative f'(x) and try to find for which values f'(x)=0, for these values of x the derivative equals zero and the tangent line is horizontal.

Example 1

Determine the minimum of the parabola:

y=2x^2-8x-4

We already know that this parabola ‘opens up’ (the coefficient of the square equals 2 and thus \geq0) and thus has a minimum that lies on the symmetry-axis, see Quadratic functions and graphs.

We differentiate the function and get:

y'=4x-8=0

and thus:

x=2

The parabola has a minimum in the point (2,-12).

Example 2

Quadratic functions (the graphs are parabolas):

y=ax^2+bx+x, (a\neq0)

have a maximum or minimum that lies on the symmetry-axis:

x_{top}=-\displaystyle\frac{b}{2a}

We can also calculate this by differentiation. The derivative of the function above is:

y'=2ax+b

The derivative is equal to 0 when:

x=-\displaystyle\frac{b}{2a}

which is the x-coordinate of the top.

The values of x for which the derivative of the function equals 0 are candidates for a maximum, minimum or horizontal inflection point, but we do not know yet which of the three is the case. This requires further investigation. There are two different methods for this.

Method 1: consider the sign of the first derivative

The derivative of a maximum in a point x=a is equal to 0.
Left to the maximum the function is increasing and thus f'(x)>0 for x<a.
Right to the maximum the function is decreasing and thus f'(x)<0 for x>a.
So the sign of the first derivative f'(x) around the maximum x=a has the following signs:

maximum : + 0 -.

Similarly, the sign of the first derivative f'(x) around the minimum x=a has the following signs:

minimum: – 0 +.

For a horizontal inflection point we find similarly:

horizontal inflection point : + 0 +   or   – 0 -.

Method 2: consider the 2nd derivative

When x=a is one of the solutions of the equation f'(x)=0 it is a candidate for a maximum, minimum or horizontal inflection point. Then the following holds.
If:

f''(a) > 0, then x=a is a minimum;

f''(a) < 0, then x=a is a maximum;

f''(a) = 0, then x=a is a horizontal inflection point.

Example 3

Check whether the function:

y=\displaystyle\frac{1}{3}x^3-x+1

has a maximum and/or a minimum.

The derivative is:

y'=x^2-1

The derivative equals 0 if:

x^2-1=0

so if:

x=1 or x=-1

The corresponding points on the graph are, respectively:

(1,\displaystyle\frac{1}{3}) or (-1,-\displaystyle\frac{5}{3})

Now we check both methods.

Method 1

Around x=-1 we find the following signs of f'x):

+ 0 – and thus there is a maximum.

Around x=1 we find the following signs of f'x):

0 + and thus there is a minimum.

Method 2

We have to calculate the second derivative:

y''=2x

Then:

y''(-1)=-2<0 and thus the function has a maximum for x=-1

and

y''(1)=2>0 and thus the function has a minimum for x=1

Example 4

Check whether the function:

y=xe^x

has a local maximum or minimum.

x.exp(x)

The derivative of the function is, using the product rule:

y'=1\cdot{e^x}+xe^x=(x+1)e^x

The derivative is 0 when x = -1 and thus the point (-1, e^{-1})  is a candidate for either a maximum or minimum. Further research should reveal which of the two it is. It is even possible that it is a point of inflection.
We can use either method, but the second method is more laborious because then we need to use the production rule again. Therefore we take method 1 and only need to consider the sign of x+1 because the exponential function e^x is always positive.

If x<-1 then x+1 is negative and if x>-1 then x+1 is positief and thus we have a minimum, see the figure.

Example 5

Investigate the function:

y=x^3

We calculate the derivative:

y'=3x^2

The equation:

3x^2=0

has x=0 as a solution which is a candidate for a maximum, minimum of horizontal inflection point.

The second derivative is:

y''=6x

which is equal for 0. So the point (0,0) is a horizontal inflection point.

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