4.2.251 AP Calculus Exam......must be true on the interval 0<x<2

In summary, the conversation discusses the integral function of $g'$ and its properties, such as always being positive and having a positive second derivative. This implies that $g$ is increasing and its graph is concave down. The conversation also mentions that $g$ is a bell-shaped function and has a point of inflection on the interval $0<x<2$.
  • #1
karush
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https://www.physicsforums.com/attachments/9527

ok from online computer I got this

$\displaystyle\int_0^x e^{-t^2}=\frac{\sqrt{\pi }}{2}\text{erf}\left(t\right)+C$

not sure what erf(t) means
 
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  • #2
\(\displaystyle \mathrm{erf(z)}\) is so called error function.

But you shouldn't need to know that to be able to solve the problem.
 
  • #3
ok but isn't this just going to be $y=\dfrac{\pi}{2}$ which is just a horizontal line
 
  • #4
karush said:
ok but isn't this just going to be $y=\dfrac{\pi}{2}$ which is just a horizontal line

the integral function defining $g’$ is not a constant

the graph of $e^{-t^2}$ is a bell shaped curve, symmetrical to the y-axis and always positive.

for $0<x<2$, $g’ > 0$ (why?)

also, $g’’ > 0$ (why again?)

... what does that say about the behavior of $g$?
 
  • #5
Karush, you are making this much harder than it should be!

You are given that [tex]y'= \int_0^x e^{-t^2}dt[/tex].

Since [tex]e^x[/tex] is positive for all x certainly both [tex]e^{-t^2}[/tex] and [tex]\int_0^x e^{-t^2}dt[/tex] are positive for all x and t. The derivative and second derivative of y are positive for all x. That is all you need.
 
  • #6
HallsofIvy said:
Karush, you are making this much harder than it should be!

You are given that [tex]y'= \int_0^x e^{-t^2}dt[/tex].

Since [tex]e^x[/tex] is positive for all x certainly both [tex]e^{-t^2}[/tex] and [tex]\int_0^x e^{-t^2}dt[/tex] are positive for all x and t. The derivative and second derivative of y are positive for all x. That is all you need.

$g’’ = e^{-x^2} > 0$ for all $x$

$g’ > 0$ for $x > 0$, $g’=0$ at $x=0$

$g’ < 0$ for $x < 0$
 

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  • #7
skeeter said:
$g’’ = e^{-x^2} > 0$ for all $x$$g’ > 0$ for $x > 0$, $g’=0$ at $x=0$
$g’ < 0$ for $x < 0$
Let g be a function with first derivative given by
$$\displaystyle g'=\int_0^x e^{-t^2}\, dt$$
Which of the following must be true on the interval $0<x<2$
a. g is increasing, and the graph of g is concave up.
b. g is increasing, and the graph of g is concave down.
c. g is decreasing, and the graph of g is concave up.
d. g is decreasing, and the graph of g is concave
e. g is decreasing, and the graph of g has a point of inflection on $0<x<2$

ok not sure which of options fit (e) looks plausible
 
  • #8
You are not sure? Haven't you read the responses here?
You are given that $g'(x)= \int_0^x e^{-t^2}dt$ and you have been told that this integral is positive for all x. (Because the integrand is positive for all t.) If a function has positive derivative is it increasing or decreasing?

By the "fundamental theorem of Calculus" [tex]g''= e^{-t^2}[/tex] which is always positive. If a function has positive second derivative is it convex upward or downward?
 
  • #9
Country Boy said:
You are not sure? Haven't you read the responses here?
You are given that $g'(x)= \int_0^x e^{-t^2}dt$ and you have been told that this integral is positive for all x. (Because the integrand is positive for all t.) If a function has positive derivative is it increasing or decreasing?

By the "fundamental theorem of Calculus" [tex]g''= e^{-t^2}[/tex] which is always positive. If a function has positive second derivative is it convex upward or downward?
g is bell shaped so it convex down
 

FAQ: 4.2.251 AP Calculus Exam......must be true on the interval 0<x<2

1. What is the significance of the interval 0

The interval 0

2. How does this statement relate to the Fundamental Theorem of Calculus?

The statement is a specific application of the Fundamental Theorem of Calculus, which states that the derivative and integral of a function are inverse operations. In this case, the statement must be true in order for the derivative and integral of the function to be equal on the given interval.

3. Can you give an example of a statement that would satisfy the given conditions?

An example of a statement that would satisfy the given conditions is "The derivative of the function f(x) is equal to the integral of the function g(x) on the interval 0

4. How does this concept apply to real-world problems?

The concept of the statement being true on a given interval is important in solving real-world problems that involve finding the derivative and integral of a function. It ensures that the solution is valid and accurate for the specific interval of interest.

5. What are some common mistakes students make when dealing with statements that must be true on a given interval?

Some common mistakes students make include forgetting to specify the interval, using the wrong interval, or incorrectly applying the Fundamental Theorem of Calculus. It is important to carefully read and understand the given conditions and to double-check the validity of the statement on the specified interval.

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