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

[karush]

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

 

[Theia]

$$\mathrm{erf(z)}$$ is so called error function.

But you shouldn't need to know that to be able to solve the problem.

 

[karush]

ok but isn't this just going to be $y=\dfrac{\pi}{2}$ which is just a horizontal line

 

[skeeter]

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 $g$?

 

[HallsofIvy]

Karush, you are making this much harder than it should be!

You are given that y&#039;= \int_0^x e^{-t^2}dt.

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

 

[skeeter]

Karush, you are making this much harder than it should be!

You are given that y&#039;= \int_0^x e^{-t^2}dt.

Since e^x is positive for all x certainly both e^{-t^2} and \int_0^x e^{-t^2}dt 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 $x < 0$

 

[karush]

$g’’ = e^{-x^2} > 0$ 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

 

[HOI]

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" g&#039;&#039;= e^{-t^2} which is always positive. If a function has positive second derivative is it convex upward or downward?

 

[karush]

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" g&#039;&#039;= e^{-t^2} 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