Determine Increasing/Decreasing and Concavity

Question

Given the function f(x)=x^3−48x^2, determine all intervals on which f is both increasing and concave up.

I am also confused about when my teacher tested values and stated whether or not a number was positive or negative using a number line.

Answer 1

What your teacher probably did is point out that f is increasing on intervals where f'(x) is positive, and concave up where f''(x) is positive. So we're looking for when

f'(x) = 3x^2-96x,

f''(x) = 6x-96 = 6(x-16)

are positive. Where the number line comes in is by noticing that f'(x) = 3x^2-96x = (3x-96)x is 0 at two points:

when x = 0 and x = 32. That means that between 0 and 32, f'(x) has to have the same sign (otherwise, since f'(x) is continuous, we'd have to cross over another solution to f'(x) = 0). So you can divide the real number line into 3 parts:

x < 0, 0 < x < 32, and x > 32, and on each of the regions, f'(x) must have the same sign on the whole region. Since f'(x) has to have the same sign on these regions, we can just select a single value from each. For example, x = -1, x = 1, x = 33.

Then plugging in, we find f(-1) = 99, f(1) = -93, f(33) = 99.

We can do the same trick with f''(x) = 6(x-16) = 0, which splits the number line into the regions

x < 16 and x >16.

This should be enough for you to finish from here.

Answer 2

Hi Brian,

A function f(x) is increasing when

df/dx >0 (1)

The graph of a function is concave up when

d²f/dx²<0 (2)

For f(x)=x^3−48x^2

df/dx=3x²-96x=3x(x-32) (3)

d²f/dx²=6x-96 (4)

From (4)

d²f/dx²<0 is true only for x<96/6=16 (5)

The condition df/dx=3x(x-32)>0 fulfills for

. x<0 or for x>32

B/c from (5) x<16 <32 ,

the only option is x<0

Hope this is helpful

Best,

Dr.Ariel B.

About your 2nd question: the convention is for points to the right of point zero to represent POSITIVE numbers , and to the left - NEGATIVE

Dr.B.

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Answer 3

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