Question

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.

Answer 1

a) the intervals where f is increasing is
`(-\infty,-1) \cup((1)/(2),\infty)`

the intervals where f is decreasing is
`(-1,(1)/(2))`

b)
`f(-1) = 6`: local maximum

`f((1)/(2)) = (-3)/(4)`: local minimum

c) The inflection point :
`x = 0.25`

the range of concavity is:
`(-\infty,0.25)`

Explanation:

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.

`f(x) = 4x^3 +3x^2 -6x +4`

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
`f'(x) = 0`

`f(x) = 4x^3 +3x^2 -6x +4`

`f'(x) = 12x^2 +6x -6`

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

`0 = 12x^2 +6x -6`

now solve for x

`0 = (2x-1)(x+1)`

`x = 1/2,\,\, x=-1`

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

`f(x) = 4x^3 +3x^2 -6x +4`

`f(-1) = 4(-1)^3 +3(-1)^2 -6(-1) +4 = 6` local maximum

`f((1)/(2)) = 4((1)/(2))^3 +3((1)/(2))^2 -6((1)/(2)) +4 = (-3)/(4)` 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
`(-\infty,-1) \cup((1)/(2),\infty)`

the intervals where f is decreasing is
`(-1,(1)/(2))`

Concavity and Inflection Points

Now, we can go further in by differentiating our
`f'(x)`

`f'(x) = 12x^2 +6x -6`

`f''(x) = 24x +6`

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

`f''(-1) = -18`

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

`f''((1)/(2)) = 18`

this this is a positive value, it shows that at this value of x, the curve looks like this:
`\cup` (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
`\cap` shape and for what range of x the curve maintains
`\cup` shape?)

we know that for the intervals
`(-\infty,-1) \cup((1)/(2),\infty)` the curve is increasing, but the shape remains like
`\cap` 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:
`f''(x) = 0`

`f''(x) = 24x +6`

`0 = 24x +6`

`x = -0.25`

this is the point where concave turns to convex.

the inflection point is:
`x = 0.25`

the range of concavity is:
`(-\infty,0.25)`

for fun we can also find the range of convexity

the range of convexity is:
`(0.25, \infty)`

hopefully, this was helpful and a fun read.