Question

How do you find and classify the critical points of `f(x) = log(x^2+x)^2`?

Answer 1

There is a single local maximum of `-ln(16)/(lna)` occurring when `x=-1/2` where `a` is the logarithm base

Explanation:

The logarithm base is not specified, so let us assume base`a`, where `a` is constant. (`a=10` for standard logarithms, and `a=e` for Natural (Napier) Logarithms).

Then, we have:

`f(x) = log_a((x^2+x)^2)`

Not that we could use the log rule:

`log a^b = bloga`

And as we wish to differentiate we should change base to Natural (base `e`) logarithms:

`f(x) = (ln((x^2+x)^2))/(lna)`

Then we can differentiate wrt `x`by applying the chain rule, to get:

`f'(x) = 1/(lna) * 1/(x^2+x)^2 * 2(x^2+x)(2x+1)`
`\ \ \ \ \ \ \ \ \ = 2/(lna) * (2x+1)/(x^2+x)`

At a critical point we require that the first derivative vanish, this requires that:

`f'(x) = 0 => (2x+1)/(x^2+x)= 0`
`:. 2x+1=0`
`:. x=-1/2`

And when `x=-1/2` we gte:

`f(-1/2) = (ln((1/4-1/2)^2))/(lna)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (ln((1/4-1/2)^2))/(lna)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (ln(1/16))/(lna)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -ln(16)/(lna)`

We can perform a second derivative test to determine the nature of the critical point:

If `f'(x) = 0` then `f''(x) \ { (lt 0,"local maximum"), (=0, "Point of Inflection"), (gt 0, "local minimum") :}`

And so we differentiate `f'(x)` wrt `x` to get:

`f''(x) = 2/(lna) { ( (x^2+x)(2) - (2x+1)(2x+1) ) / (x^2+x)^2 }`
`\ \ \ \ \ \ \ \ \ \ = 2/(lna) { ( 2x^2+2x -4x^2-4x-1 ) / (x^2+x)^2 }`
`\ \ \ \ \ \ \ \ \ \ = -2/(lna) { ( 2x^2+2x+1 ) / (x^2+x)^2 }`

And when `x=-1/2 => f''(x) < 0` and we can conclude that `x=-1/2` corresponds to a local maximum.

We should also consider the possibility of any inflection points by seeing if the second derivative vanishes:

`f''(x) =0= > 2x^2+2x+1 = 0`

Which has no real solutions.

Hence, we can conclude that there is a single local maximum of `-ln(16)/(lna)` occurring when `x=-1/2` where `a` is the logarithm base

graph{log((x^2+x)^2) [-5, 5, -4, 4]}