Given a function f(x) = 4x^3 + 3x^2 − 6x + 1 (a) Find the intervals on which f is increasing or decreasing. (b) Find the local maximum and minimum values of f. (c) Find the intervals of concavity and the inflection points.

This is a positive cubic function, (positive only means that the sign of the highest power term is +ve, nothing too fancy, its just to visualize the shape of the curve)

the positive sign tells you that this curve is coming from negative y to positive y when looking from left to right.

[tex]f(x) = 4x^3 +3x^2 -6x +4[/tex]

we can find the intervals where it is increasing and decreasing by knowing where this function has its stationary points (or turning points), in other words where [tex]f'(x) = 0[/tex]

[tex]f(x) = 4x^3 +3x^2 -6x +4[/tex]

[tex]f'(x) = 12x^2 +6x -6[/tex]

this is the function's first derivative. To find the stationary values, set [tex]f'(x) = 0[/tex]

[tex]0 = 12x^2 +6x -6[/tex]

now solve for x

[tex]0 = (2x-1)(x+1)[/tex]

[tex]x = 1/2,\,\, x=-1[/tex]

we can find the local minimum and maximum values of f by plugging in these value in the original function f(x):

[tex]f(x) = 4x^3 +3x^2 -6x +4[/tex]

[tex]f(-1) = 4(-1)^3 +3(-1)^2 -6(-1) +4 = 6[/tex] local maximum

[tex]f(\dfrac{1}{2}) = 4(\dfrac{1}{2})^3 +3(\dfrac{1}{2})^2 -6(\dfrac{1}{2}) +4 = \dfrac{-3}{4}[/tex] local minimum

We also have enough information to show the intervals at which f(x) is increasing or decreasing

Since this is a positive cubic curve, the plot is coming up from negative infinity of the y-axis all the way upto x= -1, then turns back down until it reaches x=1/2, then finally turns up again to positive infinity of the y-axis.

so,

the intervals where f is increasing is [tex](-\infty,-1) \cup(\dfrac{1}{2},\infty)[/tex]

the intervals where f is decreasing is [tex](-1,\dfrac{1}{2})[/tex]

Concavity and Inflection Points

Now, we can go further in by differentiating our [tex]f'(x)[/tex]

[tex]f'(x) = 12x^2 +6x -6[/tex]

[tex]f''(x) = 24x +6[/tex]

we can put the values we obtained of x from [tex]f'(x) = 0[/tex] to find the curvature(or shape) of the curve at those points

[tex]f''(-1) = -18 [/tex]

this is a negative value, it shows that at this value of x, the curve looks like this: [tex]\cap[/tex] (this is known a concave shape)

[tex]f''(\dfrac{1}{2}) = 18 [/tex]

this this is a positive value, it shows that at this value of x, the curve looks like this: [tex]\cup[/tex] (this is knows as the convex shape)

with this much information we have some idea about the concavity. (i.e, for what range of x does the curve maintain [tex]\cap[/tex] shape and for what range of x the curve maintains [tex]\cup[/tex] shape?)

we know that for the intervals [tex](-\infty,-1) \cup(\dfrac{1}{2},\infty)[/tex] the curve is increasing, but the shape remains like [tex]\cap[/tex] even after this range.

So what we need is a point where the two shapes begin to change:

and that is the inflection point:

to put in terms of math: the inflection point is where: [tex]f''(x) = 0[/tex]

[tex]f''(x) = 24x +6[/tex]

[tex]0 = 24x +6[/tex]

[tex]x = -0.25 [/tex]

this is the point where concave turns to convex.

the inflection point is: [tex]x = 0.25[/tex]

the range of concavity is: [tex](-\infty,0.25)[/tex]

for fun we can also find the range of convexity

the range of convexity is: [tex](0.25, \infty)[/tex]

hopefully, this was helpful and a fun read.