f(x)=x^3-4x^2+2, which of the following statements are true: (1) Increasing in (-infty, 0), decreasing in (83, +infty). (2) Increasing in both (-infty, 0), and (83, +infty). (3) decreasing in both (-infty, 0), and (83,+infty). (4) Decreasing in (-infty, 0), Increasing in (83, +infty). (5) None of the above.

solvarmedw 2022-10-02 Answered
Increasing and decreasing intervals of a function
f ( x ) = x 3 4 x 2 + 2, which of the following statements are true:
(1) Increasing in ( , 0 ), decreasing in ( 8 3 , + ).
(2) Increasing in both ( , 0 ), decreasing in ( 8 3 , + ).
(3) decreasing in both ( , 0 ), and ( 8 3 , + ).
(4) Decreasing in ( , 0 ), Increasing in ( 8 3 , + ).
(5) None of the above.
f ( x ) = 0 = 3 x 2 8 x = 0 x = 8 3 , x = 0 are the singular point/point of inflection.Could anyone tell me what next?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Paige Paul
Answered 2022-10-03 Author has 11 answers
Step 1
So, when you put f ( x ) = 0 and got x = 0 , 8 / 3, it means that function takes the “u-turn” at those points. Now, we need to check what was happening before x = 0, what’s happening between 0 and 8/3 and what will happen after x = 8 / 3.
We can take any two x’s such that x < 0 and we will find that if x 1 < x 2 then f ( x 1 ) < f ( x 2 ) or we can see that f ( x ) = 3 x 2 8 x is positive for any x < 0 and hence function is increasing in the interval ( , 0 ].
Step 2
And as we know that the function will take a “u-turn” at x = 0 so, the function will decrease between 0 and 8/3 and again after a “u-turn” at x = 8 / 3 it will increase. If you find this “u-turn” concept informal you can go for similar method above and, you will find that f is decreasing between x = 0 and x = 8 / 3, and finally it is increasing for x > 8 / 3
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-08-13
The effect of changing sample sizes on outliers
I know that the size of a sample is inversely proportional to the width of a confidence interval, and that outliers tend to increase the width of the interval as well. So that must mean that increasing the sample size reduces the effect of outliers on a confidence interval, and decreasing the sample size amplifies the effect, correct?
How can I show this using formulae instead of words for, say a confidence interval for a one-sample z-test?
Also as a side note, does changing the sample size change how outliers affect the p-value of a hypothesis test? I'm inclined to say yes, but I'm not sure how to justify that conclusion.
asked 2022-08-22
Some doubts with the sign of a derivative
Good evening to everyone. The derivative is defined in the following order:
d d x f ( x ) = x 2 x + 11 ( x + 3 ) 2 e 2 x for x < 3
d d x f ( x ) = x 2 + x 11 ( x + 3 ) 2 e 2 x for 3 < x < 2 and d d x f ( x ) = x 2 + x + 1 ( x + 3 ) 2 e x 2 for x > 2. If you'll do the sign of the first one it'll be x 2 + x 11 < 0 therefore only x 1 = 1 + 45 2 verifies the conditions, for the second one it is x 2 + x 11 > 0 therefore x 1 = 1 + 45 2 and x 1 = 1 45 2 , the last one x 2 + x + 1 > 0 has the solutions x 1 = 1 i 3 2 and x 2 = 1 + i 3 2 (here I think that I did a mistake). Therefore our function should increase on the interval ( , 1 45 2 ) then decrease on the interval ( 1 45 2 , 1 + 45 2 ) then increase again on the interval ( 1 + 45 2 , 1 i 3 2 ) and decrease again on ( 1 + i 3 2 , ). But Instead the solutions of the third equation are x 1 = 1 5 2 and x 2 = 1 + 5 2 . And it is decreasing on ( , 1 45 2 ), then increasing on ( 1 45 2 , 3 ) then decreasing on (-3,2) then increasing again on ( 2 , ). I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.
asked 2022-10-20
Question about finding where the function increases and decreases on f ( x ) = 1 x .
f ( x ) = 1 x , x 1
I have been staring at this equation for a bit. Things I'm confused on.
the derivative of this is: f ( x ) = 1 x 2 now, how am I supposed to find where this derivative increases/decreases? Do I find the critical points first? by setting the derivative to 0? or do I solve it like 1 x 2 > 0 cross multiply to make it: 1 > x 2 and if so once I square this does it make the result x = 1 , x = 1? I'm really lost here and it seems like it should be easier.
Does setting the derivative to > or < or = and solving for the x give a critical point?
asked 2022-10-06
Calculus, solving for increasing/decreasing and concavity
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
f ( x ) = ln ( x ) x
Find intervals of concavity for the graph of the function f ( x ) = ln ( x ) x .
I have already found the first and second derivative but I am confused on how to solve for it f ( x ) = ln ( x ) 1 x 2
f ( x ) = 2 ln ( x ) 3 x 3
asked 2022-08-13
If h has positive derivative and φ is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and φ is continuous and positive. If:
f ( x ) = h ( 0 x 4 4 x 2 2 φ ( t ) d t ) .
Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:
f ( x ) = h ( 0 x 4 4 x 2 2 φ ( t ) d t ) φ ( x 4 4 x 2 2 )
We need to analyze where is positive and negative. So I solved the inequalities:
x 4 4 x 2 2 > 0 x 4 4 x 2 2 < 0
That gives: ( , 2 ) ( 2 , ) for the first case and ( 2 , 2 ) for the second one. Then (not sure of this part) h ( 0 x 4 4 x 2 2 φ ( t ) d t ) > 0 and φ ( x 4 4 x 2 2 ) > 0 if x ( , 2 ) ( 2 , ) . Also if both h′ and φ are negative the product is positive, that's for x ( 2 , 2 ).
The case of the product being negative implies:
x [ ( , 2 ) ( 2 , ) ] ( 2 , 2 ) = [ ( , 2 ) ( 2 , 2 ) ] [ ( 2 , ) ( 2 , 2 ) ] = .
So the function is increasing in ( , 2 ) , ( 2 , 2 ) , ( 2 , ). So the function does not have maximum or minimum. Not sure of this but what do you think?
asked 2022-08-11
Is a Square Bracket Used in Intervals of Increase/Decrease?
For example, the I.O.I of y = x 2 is (0,infinite), with the round brackets meaning that the value is excluded. Are there any scenarios where a square bracket would be used when stating the intervals of increase/ decrease for a function? If it narrows it down, the only functions I deal with are: linear, exponential, quadratic, root, reciprocal, sinusoidal, and absolute
asked 2022-08-12
Increasing and decreasing intervals
Find a polynomial f(x) of degree 4 which increases in the intervals ( , 1 ) and (2,3) and decreases in the intervals (1,2) and ( 3 , ) and satisfies the condition f ( 0 ) = 1.
It is evident that the function should be f ( x ) = a x 4 + b x 3 + c x 2 + d x + 1. I differentiated it. Now, I'm lost. I tried putting f ( 1 ) > 0, f ( 3 ) f ( 2 ) 0, and f ( 2 ) f ( 1 ) 0. Am I doing correct? Or is there another method?