# Cubic Functions: Derivatives and Graph Tendencies

• chubbyorphan
In summary, the conversation is about a student struggling with a calculus problem involving derivatives and the behavior of a graph. They are specifically trying to find intervals of increase or decrease and points of inflection, but are having trouble with the math. Another user helps clarify some misunderstandings and provides guidance on how to approach the problem. The student eventually realizes their mistake and thanks the other user for their help.
chubbyorphan

## Homework Statement

hey Forum! I had a question here I'm struggling with and was wondering if someone could take a look. its Dealing with calculus, specifically derivatives and behaviors of the graph:

http://i41.tinypic.com/mc6opj.jpg

I just started and part a) already has me stumped D:

## The Attempt at a Solution

for intervals of increase or decrease I know we must equate the first derivative to zero.

f(x) = x^3 – x^2 + 4x – 3
f’(x) = 3x^2 – 2x + 4
f’’(x) = 6x – 2

so:
f’(x) = 3x^2 – 2x + 4
0 = 3x^2 – 2x + 4
however I couldn't factor this.. and when I tried to use the quadratic formula.. I got a negative discriminant.
From what I gather this means the derivative has no real roots.. However the actual graph of the original function crosses the x-axis just before 1.. I'm confused! how do I calculate this?

any advice would be wicked!

chubbyorphan said:

## Homework Statement

hey Forum! I had a question here I'm struggling with and was wondering if someone could take a look. its Dealing with calculus, specifically derivatives and behaviors of the graph:

http://i41.tinypic.com/mc6opj.jpg

I just started and part a) already has me stumped D:

## The Attempt at a Solution

for intervals of increase or decrease I know we must equate the first derivative to zero.

f(x) = x^3 – x^2 + 4x – 3
f’(x) = 3x^2 – 2x + 4
f’’(x) = 6x – 2

so:
f’(x) = 3x^2 – 2x + 4
0 = 3x^2 – 2x + 4
however I couldn't factor this.. and when I tried to use the quadratic formula.. I got a negative discriminant.
From what I gather this means the derivative has no real roots.. However the actual graph of the original function crosses the x-axis just before 1.. I'm confused! how do I calculate this?

any advice would be wicked!
Since f'(x) isn't 0 for any real number x, then there are no points on the cubic's graph with a horizontal tangent. Since the derivative is never zero, it must always be positive or always negative, meaning that the original function is either always increasing or always decreasing.

Thanks! okay that makes a lot of sense actually :D
SO its one or the other, right?

How do I prove whether its constantly increasing or decreasing using derivatives?

cuz normally I would equate the derivative to zero.. take the values that gives me for x..
and consider those 'significant points'.. then I would check the value for the derivative at points after, before, and possibly in between these 'signicant points' to determine where the original function is increasing or decreasing.

but since I have no real values that equate the derivative to zero.. I have no significant points to work with..

and the function crosses the x-axis just before x = 1 (according to my graphing calculator)

here is a picture:
http://i41.tinypic.com/bgoaah.jpg

but based on this.. am I correct to say:
there is no minimum or maximum?
We can attest to this simply because the function does is solely ^increasing? is that right?

How do I calculate the point in which the function changes from concave down to concave up, or the point of inflection?
Nvm. I think I got point of inflection.. equate second derivative to zero!
then sub x value into original function! cha ching :D

I know I'm asking a lot but this graph really has me grinding my teeth.. thank you so much for helping me through this!

You seem to be confusing f(x)=0 and f'(x)=0. If f'(x) is never zero, then to see whether the function is increasing or decreasing you just evaluate f'(x) at any point (just like you would if you had special critical points, except you only have to check f'(x) at one point instead of a couple)- in this example, f'(0)=4 so the function f(x) is always increasing. The fact that f(x)=0 when x is approximately 1 is no big deal

Aha! Office_Shredder, I see what you mean! you're right, I was jumbling things up there a little bit! Thanks for clearing that up, homie!

## What is a cubic function?

A cubic function is a polynomial function of degree 3, which means it has the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants.

## What is a derivative?

A derivative is a measure of how a function changes as its input value changes. It is represented by the symbol f'(x) and can be interpreted as the slope of the tangent line to the function at a specific point.

## How do you find the derivative of a cubic function?

To find the derivative of a cubic function, you can use the power rule, which states that the derivative of a function of the form f(x) = x^n is f'(x) = nx^(n-1). In the case of a cubic function, the derivative would be f'(x) = 3ax^2 + 2bx + c.

## What is the relationship between a cubic function and its derivative graph?

The derivative graph of a cubic function is a quadratic function, which means it has the form f(x) = px^2 + qx + r. The derivative graph shares the same x-intercepts as the original cubic function, and the slope of the derivative graph at a specific point corresponds to the concavity of the original cubic function at that point.

## What are the key features of a cubic function's graph?

The key features of a cubic function's graph include its y-intercept (d), its end behavior (determined by the leading coefficient a), its critical points (where the derivative is equal to 0), and its inflection points (where the concavity changes).

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