Question

Let g be the function given by g(x) integral 0 to x sin(t^2)dt for -1 less than or equal to x and less than or equal to 3. On which of the following intervals is g decreasing?

Answer 1

This is your g(x), correct?
\[g(x)=\int\limits_{0}^{x}\sin(t^2)dt\]

Answer 2

\[\int\limits_{0}^{x}\sin (t^{2}) dt for -1\le x \le 3\]

Answer 3

yes

Answer 4

g(x) is decreasing if g'(x)=0. What is g'(x)?

Answer 5

i dont know how to do g'(x) for this problem

Answer 6

Use the Fundamental Theorem of Calculus part I which states that if:
\[g(x)=\int\limits_{a}^{x}f(t) dt\]
for some function f(x), then g'(x)=f(x), provided f is continuous and a is any constant.
Since f(t)=sin(t^2) in this case, g'(x)=sin(x^2).
Can you finish it now?

Answer 7

no im not getting an answer

Answer 8

d[g(x)]/dx = f(x)
function is decresing were it's derivative is <0. So where f(x)<0

Answer 9

sin(x^2)=0 for x^2=k*pi and thus x=sqrt(2k*pi). For k=0, we have x=0. For k=1, x=sqrt(pi). k=2: x=sqrt(2pi)>3. Finally, since the graph of sin(x^2) is symmetrical about the y-axis, we divide the interval [-1,3] into 3 parts: [-1,0],[0,sqrt(pi)],[sqrt(pi),sqrt(2pi)],[sqrt(2pi),3] and we find out that g(x)<0 on the third interval. Thus, f(x) is decreasing there.

Answer 10

sin(x^2)<0
find the interval where this is true

Answer 11

sin(x^2)=0 is a extremum point. In that point function is not incresing, and not decresing. It's stationary....

Answer 12

Oh, so the answer should be (sqrt(pi),sqrt(2pi)). My bad. :))

Answer 13

thes are the choices
a.) -1 less than or equal to x and less than or equal to 0
b.) 0 less than or equal to x and less than or equal to 1.772
c.) 1.253 less than or equal to x and less than or equal to 2.171
d.) 1.772 less than or equal to x and less than or equal to 2.507
e.) 2.802 less than or equal to x and less thanor equaL to 3