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

Answer:

a) the intervals where f is increasing is $(-\infty,-1) \cup(\dfrac{1}{2},\infty)$

the intervals where f is decreasing is $(-1,\dfrac{1}{2})$

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

$f(\dfrac{1}{2}) = \dfrac{-3}{4}$: local minimum

c) The inflection point :  $x = 0.25$

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

Step-by-step 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

$f(\dfrac{1}{2}) = 4(\dfrac{1}{2})^3 +3(\dfrac{1}{2})^2 -6(\dfrac{1}{2}) +4 = \dfrac{-3}{4}$ 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 $(-\infty,-1) \cup(\dfrac{1}{2},\infty)$

the intervals where f is decreasing is $(-1,\dfrac{1}{2})$

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)

$f''(\dfrac{1}{2}) = 18$

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 $(-\infty,-1) \cup(\dfrac{1}{2},\infty)$ 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.