Walk through please.

Question

1.Find and classify (categorize the point as maximum, minimum or inflection point) of the function below. 

f(x)= -x^3-3x^2+1



2.Find the critical points below
f(x)=x^3+3x^2+1


3. Find and classify (categorize the point as maximum, minimum or inflection point) of the function below. 
f(x)=x^5


4. Determine the concavity and relative extrema of the function below.
f(t)=x^4-32x+56

Answer 1

Eva,

I'll do my best, but this is a lot to "walk through" without diagrams. So to begin with, 1, 2, and 3 are all the same question about different functions. Critical points are the local minimums, local maximums, and inflection points. And the local minimums and maximums are also known as relative extrema, as requested in number 4.

 

Minimums and maximums can be identified as points where the slope of the curve is 0, i.e. f'(x) = 0.

Inflection points, likewise, are found where f"(x) = 0. So let's walk through number 1.

 

1) f(x) = x³ - 3x² + 1     so       f'(x) = 3x² - 6x    and    f"(x) = 6x - 6

 

3x² - 6x = 0          set f' = 0

3x(x - 2) = 0         factor

3x = 0    x-2 = 0    zero product rule

x = 0     x = 2

 

Plug these values into the original equation to find the y-value.

f(0) = 1      f(2) = 8 - 12 + 1 = -3

giving us the points (0,1) and (2,-3).  It seems like you should be done with high/low at this point, but it is important to verify the slope trends around these points. To do so, we pick test values of x to plug into the first derivative.

f'(-1) = 3 + 6 = 9 or positive slope to the left of 0

f'(1) = 3 - 6 = -3  or negative slope between 0 and 2

f'(3) = 27 - 18 = 9 or positive slope to the right of 2

 

We have now verified that (0,1) is a local high and (2,-3) is a local low.

We use the term local because they are not absolute. Odd degree functions like this follow a general trend from negative infinity to positive infinity, or vice versa. The absolute extremes are undefined.

 

6x - 6 = 0   to find inflection point

6x = 6

x = 1         f(1) = 1 - 3 + 1 = -1      (1,-1) is an inflection point.

 

2) follow the same exact steps to find (-2,5) local high

                                                       (0,1) local low

                                                       (-1,3) inflection point

 

3) This is an example of why we need the slope test around the f'=0 values.

 

f(x) = x⁵        f'(x) = 5x⁴        f"(x) = 20x³

 

5x⁴ = 0                               20x³ = 0

x = 0                                      x = 0

 

With a degree 5 equation like this, we typically expect to see 4 relative extrema. But we only see one. It also turns out to be an inflection point. If we check the slopes around x = 0,

 

f'(-1) = 5        f'(1) = 5    BOTH POSITIVE

 

What we have here is a case where there is no local highs or lows, only a momentary leveling off on a continuously upward path. And that occurs at the inflection point (0,0).

 

4) The are two things different about number four. First, because the degree is even (4), there will be an absolute extreme. Second, they have asked for concavity. The inflection points will give us the divisions between positive and negative concavity, but we must again use test values to identify which zones are positive (concave up) and which are negative (concave down).

 

f(x) = x⁴ - 32x + 56         f'(x) = 4x³ -32            f"(x) = 12x²

 

4x³ -32 = 0

4x³ = 32

x³ = 8

x = 2

 

Because this is even degreed and positive in the leading coefficient, we know that this is an absolute low.

And because it is the only zero slope, there are no local extrema.

f(2) = 16 - 64 + 56 = 9        (2,8) is absolute minimum.

 

12x² = 0

x = 0               f(0) = 56     (0,56) is an inflection point

 

f"(-1) = 12      f"(1) = 12   Both positive. So the function is entirely concave up except for the momentary hiccup of 0 at (0,56).

 

I hope that helps.