Question

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

Answer 1

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.