MA511 Test 2 Practice

f(x) = integral from -10 to x^2 + 2x e^t^2 dt Find the value of x where f(x) attains its minimum. (Hint: you will need the Chain Rule.)

-1

Let f(x)=4x−1, and suppose ϵ=0.2. Find δ so that |x−1|=δ⇒|f(x)−f(1)|=ϵ. Give an exact decimal answer.

.05

Compute limx→0 (xsin(1x))

0

Consider the Dirichlet function on the domain [0,1] f(x)= { 1 if x is rational and 0 if not Let P={0,.1,.3,.45,.7,12‾√,.9,1} Compute L(f,P) and U(f,P)

L(f,P) = 0 U(f,P) = 1 Dirichlet function is not Reimann Integrable

Let D:R->R be the Dirichlet function D(x) = {1 if x rational and 0 otherwise Suppose f(x) = xD(x) Where on the domain is f continuous?

0 only

Suppose that f is continuously differentiable [=f′ exists and is continuous] on [1,4] with f(1)=2. Suppose also that |f′(x)|≤0.3 for all x∈[1,4]. Give the best bounds we can for f(4) (as a decimal):

1.1 <= f(4) <= 2.9

integrate from 0 to 1 ln(x)^2

2

integrate 1/x from -inf to inf

DNE

True or False There exists a continuous function on R whose range is Q. (Be sure to be able to explain your answer.)

False

Can you construct a function which is continuous at exactly two points? What about at exactly three points? Countably many points? Describe how you would do it.

In lecture 11, we were given h(x) = {x if x is rational and 0 if x is not rational} where h(x) is only continuous at 0. Building off of this, if we want a function that is continuous at two points, we can have j(x) = {x^2 when x is rational and 1 when x is not rational}. This function is continuous at -1 and 1. If we keep building on this, we can get continuity at three points if k(x) = {x^3 when x is rational and x when x is not rational). We can continue to build on this idea...continuous at four points m(x) = {x^2(x-2)(x+2) when x is rational and x when x is not rational}.. We can continue in this way and find functions that are continuous at countably many points.

Suppose that f:[a,b]→ℝ is a bounded Riemann integrable function. We will show that |f|:[a,b]→ℝ is also integrable.

Quiz 6 #5

Suppose that f is a continuous function on ℝ such that f(q)=0 for all q∈Q. Prove that f(x)=0 for all x∈R

Suppose for the sake of contradiction that there exists s in Q complement for which f(s)≠0. Using the density of the rational numbers in ℝ, there must exists a sequence {xn}→s for which x_n ∈ Q for all n in N. Since f is continuous at s, we must have {f(xn)}→f(s). However, notice that f(x_n)=0 for all n∈N, which implies that limf(xn)=0. As limf(x_n) is not equal to f(s), we have reached a contradiction. Thus it must be the case that f(s)=0

integrate from 0 to 1 ln(x) dx

use integration by parts -1

integrate from 3 to 6 2xe^x^2 dx

use u substitution e^36 - e^9