Question

ughh can someone please help me??

Answer 1

With what?

Answer 2

？

Answer 3

if the function g is defined as g(x) (t squared -3t -40dt on the interval [-7,5]. then g(x) has a local minimum at x=

Answer 4

as before
\[g'(x)=x^2-3t-4\] although it looks like you changed the -4 into -40
is that a typo?

Answer 5

if it is -4 then set
\[x^2-3x-4=0\] get
\[(x-4)(x+1)=0\] so your critical points are at x = -1, x = 4

Answer 6

yeah thats a typo, but are u sure i need to find the derivative? or do i just test -4, 1, -7, and 5

Answer 7

well your integral will be zero at the left hand endpoint for sure. you want the local min.

Answer 8

o ha nevermind what i just said

Answer 9

now
\[x^2-3x-4\] is a parabola that opens up, is positive outside the zeros, and negative between them. therefore -1 will be a local max and 4 will be a local min

Answer 10

or if you want to be fancy you can say
\[g''(x)=2x-3\] and
\[g''(4)=5\] and since the second derivative is positive, function is concave up, making 4 the x - value of the local min

Answer 11

first picture here shows that 4 is a local min of the function
http://www.wolframalpha.com/input/?i=integral+-7+to+x+of+t^2+-+3t-4+dt+graph

Answer 12

ok thanks a ton, this helped a lot! :)