Given g(x) is the integral from a to x of f(t) dt, we were able to derive that g'(x) = 1/h integral from x to x+h of f(t) dt. By the extreme value theorem there exists a min m and max M on [x,x+h] such that f(u) = m and f(v) = M and f(u)h<= integral x to x+h f(t) dt <= f(v)h (by properties of integrals). Dividing by h we get f(u)<= (g(x+h)- g(x)) / h <= f(v) by substitution. Here is where I'm confused, the text says if we let h--> 0 then u --> x and v--> x since u and v lie in [x,x+h]. I'm not understanding how u--> x and v--> x.

Compute the gradients of each of the following functions: o = xyz & = ln [(x² - y²)-¹/2] r = || = √√√x² + y² + x² 1/r Note: Remember that your tutorial has Hints in it, at the end of each chapter. If you're stuck, they are often quite useful! • If the answer is a scalar, you can just type it in the box (using the Calcpad if you like, or using / for fractions, ^ for exponents, shift and - for subscripts, etc.) For multiplication, you can either leave a space, or use *. So x*x=xx= x², Note that this is not the same as xx without a space; that will get read as an entirely different variable! • If you need to enter a vector, enter an ordered list of components, so for A you can enter either A = (A₂, Ay, A₂) = {A₂, Ay, A₂}. Note that the system isn't great with multiplying through by overall factors, so it's better not to write e.g. Vo= V = Vr = V= (2A, 2A, 2A₂).

1) Estimate the derivative at each point using a right hand approximation where possible 0 f(x) 5 1 2 3 11 14 22