What are the local extrema, if any, of f(x) =x^2 + 9x +1 ?

1 Answer
Mar 8, 2016

Parabolae have exactly one extrema, the vertex.

It is (-4 1/2, -19 1/4).

Since {d^2 f(x)}/dx=2 everywhere the function is concave up everywhere and this point must be a minimum.

Explanation:

You have two roots to finding the vertex of the parabola: one, use calculus to find were the derivative is zero; two, avoid calculus at all costs and just complete the square. We're going to use calculus for the practice.

f(x)=x^2+9x+1 , we need to take the derivative of this.

{d f(x)}/dx={d }/dx(x^2+9x+1)

By the linearity of the derivative we have

{d f(x)}/dx={d }/dx(x^2)+{d }/dx(9x)+{d }/dx(1).

Using the power rule, d/dx x^n = n x^{n-1} we have

{d f(x)}/dx=2*x^1+9*1*x^0+0=2x+9.

We set this equal to zero to find the critical points, the local and global minima and maxima and sometimes points of inflection have derivatives of zero.

0=2x+9 => x=-9/2,
so we have one critical point at x=-9/2 or -4 1/2.

To find the y coordinate of the critical point we sub in x=-9/2 back into the function,

f(-9/2)=(-9/2)^2+9(-9/2)+1 = 81/4 - 81/2 + 1
=81/4 - 162/4 + 4/4=-77/4=-19 1/4.

The critical point/vertex is (-4 1/2, -19 1/4).

We know that because a>0, this is a maximum.
To formally find if it's a maxima or minima we need to do the second derivative test.

{d^2 f(x)}/dx={d }/dx(2x+9)={d }/dx(2x)+{d }/dx(9)=2+0=2

The second derivative is 2 at all values of x. This means it is greater then zero everywhere, and the function is concave up everywhere (it's a parabola with a>0 after all), so the extrema must be a minimum, the vertex.