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

$\frac{dy}{dx}=\frac{2x+1}{x^2+x-20}$

Then, the condition is given by

$\frac{dy}{dx}=\frac{2x+1}{x^{2}+x-20}>0$

which implies that

$\begin{gathered} 2x+1>0 \\ then \\ x<-\frac{1}{2} \end{gathered}$

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

$(-5,-\frac{1}{2})\cup(4,\infty)$

Now, the function is decreasing when

$\frac{dy}{dx}<0$

which give us

$\begin{gathered} \frac{dy}{dx}=\frac{2x+1}{x^{2}+x-20}<0 \\ 2x+1<0 \\ then \\ x>-\frac{1}{2} \end{gathered}$

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

$(-\infty,-5)\cup(-\frac{1}{2},4)$

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

$\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{2x+1}{x^2+x-20})$

which gives

$\frac{d^2y}{dx^2}=\frac{(x^2+x-20)(2)-(2x+1)(2x+1)}{(x^2+x-20)^2}$

or equivalently

$\frac{d^2y}{dx^2}=\frac{2(x^2+x-20)-(4x^2+4x+1)}{(x^2+x-20)^2}$

which can be written as

$\frac{d^2y}{dx^2}=\frac{2x^2+2x-40-4x^2-4x-1}{(x^2+x-20)^2}$

then, we get

$\frac{d^2y}{dx^2}=\frac{-2x^2-2x-41}{(x^2+x-20)^2}$

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

$\frac{dy}{dx}=\frac{2x+1}{x^{2}+x-20}=0$

which gives

$\begin{gathered} 2x+1=0 \\ then \\ x=-\frac{1}{2} \end{gathered}$

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

$x=-\frac{1}{2}$

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,

$\frac{d^{2}y}{dx^{2}}=\frac{-2(-\frac{1}{2})^2-2(-\frac{1}{2})-41}{((-\frac{1}{2})^2+(-\frac{1}{2})-20)^2}$

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

$-2(-\frac{1}{2})^2-2(-\frac{1}{2})-41=-\frac{2}{4}+1-41=-40.5$

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

$\frac{d^2y}{dx^2}<0$

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