Question

Hi there,i am having some trouble solving the following two questions relating to extrema and intervals:

Answer 1

We have the following function:

`y=ln|x^2+x-20|`

The graph of this function is given by:

We know that a function is increasing is the first derivative is greater than zero. The derivative of the given function is given by

`(dy)/(dx)=(2x+1)/(x^2+x-20)`

Then, the condition is given by

`(dy)/(dx)=(2x+1)/(x^(2)+x-20)>0`

which implies that

`\begin{gathered} 2x+1>0 \\ then \\ x<-(1)/(2) \end{gathered}`

By means of this result and the graph from above, f(x) is increasing for x in:

`(-5,-(1)/(2))\cup(4,\infty)`

Now, the function is decreasing when

`(dy)/(dx)<0`

which give us

`\begin{gathered} (dy)/(dx)=(2x+1)/(x^(2)+x-20)<0 \\ 2x+1<0 \\ then \\ x>-(1)/(2) \end{gathered}`

Then, by means of this result and the graph from above, the function is decreasing on the interval:

`(-\infty,-5)\cup(-(1)/(2),4)`

In order to find the local extremal values, we need to find the second derivative of the given function, that is,

`(d^2y)/(dx^2)=(d)/(dx)((2x+1)/(x^2+x-20))`

which gives

`(d^2y)/(dx^2)=((x^2+x-20)(2)-(2x+1)(2x+1))/((x^2+x-20)^2)`

or equivalently

`(d^2y)/(dx^2)=(2(x^2+x-20)-(4x^2+4x+1))/((x^2+x-20)^2)`

which can be written as

`(d^2y)/(dx^2)=(2x^2+2x-40-4x^2-4x-1)/((x^2+x-20)^2)`

then, we get

`(d^2y)/(dx^2)=(-2x^2-2x-41)/((x^2+x-20)^2)`

From the above computations, the critical value point is obtained from the condition

`(dy)/(dx)=(2x+1)/(x^(2)+x-20)=0`

which gives

`\begin{gathered} 2x+1=0 \\ then \\ x=-(1)/(2) \end{gathered}`

This means that the critical point (maximum or minimum) is located at

`x=-(1)/(2)`

In order to check if this value corresponds to a maximum or mininum, we need to substitute it into the second derivative result, that is,

`(d^(2)y)/(dx^(2))=(-2(-(1)/(2))^2-2(-(1)/(2))-41)/(((-(1)/(2))^2+(-(1)/(2))-20)^2)`

The denimator will be positive because we have it is raised to the power 2, so we need to check the numerator:

`-2(-(1)/(2))^2-2(-(1)/(2))-41=-(2)/(4)+1-41=-40.5`

which is negative. This means that the second derivative evalueated at the critical point is negative:

`(d^2y)/(dx^2)<0`

which tell us that the critical value of x= -1/2 corresponds to a maximum.

Since there is only one critical point, we get:

f(x) has a local minimum at x= DNE

f(x) has a local maximum at x= -1/2