For #f(x) = (6x)/(1 + 7x^2)#, show that it changes from increasing to decreasing at the point #a# and increasing to decreasing at the point #b#, and find the minima and maxima? What is #a xx b?#

I'm guessing the question refers to the #x# coordinates where #f# changes direction, and that #a xx b = -1/7#:

#a = -sqrt(1/7)#, local minimum

#b = sqrt((1/7)#, local maximum

#f(x) = (6x)/(1 + 7x^2)#:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

#(df)/(dx) = -(6(7x^2 - 1))/(1 + 7x^2)^2#:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

#(d^2f)/(dx^2) = (84x (7x^2 - 3))/(1 + 7 x^2)^3#:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}

DISCLAIMER: LONG ANSWER!

Well, first of all, I think the question has a funky typo. It should say, "changes from decreasing to increasing at the point #a# . . . "

Whenever a function changes from increasing to decreasing or vice versa, it exhibits either a local maximum or minimum.

For example:

• #y = x^2# has one local minimum at #x = 0#, since the graph has only an upwards concavity.

graph{x^2 [-2.464, 2.464, -2, 2]}

• #y = sinx# from #0# to #2pi# changes from increasing to decreasing at #x = pi/2#, and changes from decreasing to increasing at #x = (3pi)/2#.

graph{sinx [-0.05, 6.25, -1.2, 1.01]}

For your graph, the first derivative with respect to #x#, denoted #(df)/(dx)#, shows you whether the function increases, decreases, or neither at a certain #x#.

Here, we have (using the product rule):

#(df)/(dx) = 6x cdot -1/(1 + 7x^2)^2 cdot 14x + 1/(1 + 7x^2) cdot 6#

#= (-6(14x^2))/(1 + 7x^2)^2 - (-6(1 + 7x^2))/(1 + 7x^2)^2#

#= (-6(14x^2 - (1 + 7x^2)))/(1 + 7x^2)^2#

#= -(6(7x^2 - 1))/(1 + 7x^2)^2#

So, you took the derivative correctly. Now, to find the points #a# and #b# where the function has a local maximum or minimum, find the zeroes of this derivative (i.e. find when #(df)/(dx) = 0#, the function changes direction).

#0 = -(cancel(6)^(ne 0)(7x^2 - 1))/cancel((1 + 7x^2)^2)^(ne 0)#

The denominator can never equal #0# because it's an upwards-shifted quadratic, so we don't have to worry about it.

#=> 7x^2 = 1#

So, your critical x values (for your local maxima and/or minima) are at

#=> color(green)(x) = pmsqrt(1/7) ~~ color(green)(pm 0.3780)#

Now that you know the #x# coordinates, you'll need the #y# coordinates, so plug in your #x# values back into the ORIGINAL function:

#color(green)(f(pm sqrt(1/7))) = (6(pmsqrt(1/7)))/(1 + 7(pmsqrt(1/7))^2)#

#= (pm6/(sqrt7))/(1 + 7 cdot 1/7) = pm3/(sqrt7)#

#~~ color(green)(pm 1.134)#

Right now, all you know is where the points #a# and #b# are, but not which is which.

The second derivative tells you the concavity; up, down, or an inflection point. By the question wording, one is concave up and the other is concave down.

This derivative would be somewhat ugly

#(d^2f)/(dx^2) = -d/(dx)[(42x^2 - 6)/(1 + 7x^2)^2]#,

but you should get:

#= (84x (7x^2 - 3))/(1 + 7 x^2)^3#

At the same critical #x# values you got earlier, evaluate this second derivative.

• If it is positive, the function is concave up and is at a minimum.
• If it is negative, the function is concave down and is at a maximum.
• Otherwise, the function is at an inflection point.

At #ul(x = -sqrt(1/7))#:

#ul((d^2f)/(dx^2)) = (84(-sqrt(1/7)) (7(-sqrt(1/7))^2 - 3))/(1 + 7 (-sqrt(1/7))^2)^3#

#= ((-84/sqrt7) (-2))/(2)^3 " "ul(> 0)#

At #ul(x = +sqrt(1/7))#, we similarly get:

#ul((d^2f)/(dx^2) < 0)#

So, the left-hand critical point is a minimum and the right-hand critical point is a maximum.

#color(blue)((-sqrt(1/7), -3/sqrt7) ~~ (-0.3780, -1.134))# (minimum)

#color(blue)((sqrt(1/7), 3/sqrt7) ~~ (0.3780, 1.134))# (maximum)

And we can check by graphing. Here is the function, followed by its first derivative, and then its second derivative.

#f(x) = (6x)/(1 + 7x^2)#:

graph{(6x)/(1 + 7x^2) [-3.464, 3.464, -1.732, 1.732]}

#(df)/(dx) = -(6(7x^2 - 1))/(1 + 7x^2)^2#:

graph{-(6(7x^2 - 1))/(1 + 7x^2)^2 [-3.464, 3.464, -1.732, 1.732]}

#(d^2f)/(dx^2) = (84x (7x^2 - 3))/(1 + 7 x^2)^3#:

graph{(84x (7x^2 - 3))/(1 + 7 x^2)^3 [-2.464, 2.464, -24, 24]}