Curve Sketching of f = x^2, f' = 2x

  • Thread starter Miike012
  • Start date
  • Tags
    Curve
In summary, the function f(x) = x^2 has a derivative f'(x) = 2x, and on the interval (0, ∞), the function f increases, while on the interval (-∞, 0), the function f decreases. This means that as x approaches zero from the left, the function f decreases, even though the values of y are increasing. The concept of increasing or decreasing in mathematics refers to the movement on the number line, not the magnitude of the values.
  • #1
Miike012
1,009
0
f= x^2
f' = 2x


f'> 0 in (0, ∞)... f = y increases on this interval
f' < 0 in (-∞,0) f = y decreases on this interval


I don't understand... it is saying that f decrease on (-∞,0) but obviously its not because (∞)^2 = ∞ which is increasing...

Does this mean that the function is decreasing as as x approaches zero from the left?? If it did that would make more sense to me..
 
Physics news on Phys.org
  • #2
Miike012 said:
f= x^2
f' = 2x


f'> 0 in (0, ∞)... f = y increases on this interval
f' < 0 in (-∞,0) f = y decreases on this interval


I don't understand... it is saying that f decrease on (-∞,0) but obviously its not because (∞)^2 = ∞ which is increasing...

Does this mean that the function is decreasing as as x approaches zero from the left?? If it did that would make more sense to me..

It means that if x1 < x2 < 0 then f(x1) > f(x2). For example, what is f(-10)? What is f(-5)? Is -10 less than -5?

RGV
 
  • #3
f(-10) = 100 What is f(-5) = 25... so how does that mean f is decreasing on (- inf , 0)..
But as x gets larger negative y increases... so obviously y is not decreasing on (- inf , 0).
 
  • #4
Miike012 said:
f(-10) = 100 What is f(-5) = 25... so how does that mean f is decreasing on (- inf , 0)..
But as x gets larger negative y increases... so obviously y is not decreasing on (- inf , 0).

As x *increases* from -10 to -5, f decreases from 100 to 25. That is what we mean by a decreasing function. As I have said already, f decreases if x1 < x2 gives f(x1 ) > f(x2), and that is exactly what happens for negative x values. What part of that statement do you not understand?

RGV
 
  • #5
I thought it meant that the y-values constantly decrease..

For instance if [xn , x0] where x0 <= 0

Then as x ---> xn that f(x) --> smaller and smaller y values...
Look at my picture...
 

Attachments

  • graph.jpg
    graph.jpg
    5.9 KB · Views: 424
  • #6
For negative quantities, greater magnitude means less value.
Think: when you spend money - a sum of m - it is like a negative income. When are you richer: if your income is -100 or if it is -5?

ehild
 
  • #7
Miike012 said:
I thought it meant that the y-values constantly decrease..

For instance if [xn , x0] where x0 <= 0

Then as x ---> xn that f(x) --> smaller and smaller y values...
Look at my picture...

We increase something by adding a positive quantity to it. If it starts negative, adding a positive quantity to it either makes the result "less negative" (i.e., smaller in SIZE) or else positive (which may be either smaller in size or larger in size, depending on details). Similarly, we decrease something by adding a negative amount to it, or equivalently, by subtracting a positive amount from it. If it starts negative, decreasing it makes the result "more negative" (larger in SIZE). "Increasing" means we move to the right on the real number line, while decreasing means moving to the left. It does not matter where we start from. You need to stop confusing magnitude and position. Words in mathematics have specific meanings, and these may not always agree 100% with everyday usage.

RGV
 

1. What is the equation for f(x) = x^2?

The equation for f(x) = x^2 is a quadratic function, which is represented by a parabola on a graph. The function takes the input value of x and squares it, resulting in a U-shaped curve.

2. How do you find the critical points of f(x) = x^2?

In order to find the critical points of f(x) = x^2, you must first take the derivative of the function. In this case, the derivative is f'(x) = 2x. Then, set the derivative equal to zero and solve for x. The resulting value(s) of x will be the critical points.

3. What is the significance of the slope at the critical points of f(x) = x^2?

The slope at the critical points of f(x) = x^2 is zero. This means that the tangent line at these points is parallel to the x-axis. This also indicates that the function is changing from increasing to decreasing or vice versa at these points.

4. How do you determine the concavity of f(x) = x^2?

The concavity of f(x) = x^2 is determined by the second derivative of the function, which is f''(x) = 2. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down.

5. What does the curve look like for f(x) = x^2?

The curve for f(x) = x^2 is a parabola, which is a smooth, U-shaped curve. The vertex of the parabola is located at the point (0,0) and the curve extends infinitely in both directions. The curve is symmetrical about the y-axis and the x-axis is the axis of symmetry.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
606
  • Calculus and Beyond Homework Help
Replies
8
Views
470
  • Calculus and Beyond Homework Help
Replies
1
Views
463
  • Calculus and Beyond Homework Help
Replies
3
Views
286
  • Calculus and Beyond Homework Help
Replies
2
Views
543
  • Calculus and Beyond Homework Help
Replies
9
Views
546
  • Calculus and Beyond Homework Help
Replies
1
Views
282
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
887
Back
Top