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

1 Answer
Mar 8, 2016

Answer:

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.