Theorums
Derivatives
Antiderivates
Word problems
100

What does the Extreme Value Theorem guarantee about a function that is continuous on a closed interval [a,b]?

The function has both an absolute maximum and an absolute minimum on that interval.

100

y=ln(cscx - cotx)

-cscxcotx + csc²x / cscx - cotx

100

Find an antiderivative of the function f(x)=2x+5.

x+5x+C

100

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s. When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

140 cm2/s

200

What is the condition on a function f that must be met to apply the Mean Value Theorem on an interval [a,b]?

The function must be continuous on the closed interval [a,b] and differentiable on the open interval (a,b).

200

g(x) = log3(23- 4x²+5)

g'(x)= 3x²-8x / (x3-4x²+ 5) (ln (3)) 

200

Find the antiderivative of f(x)=3x2+2x−5.

x3+x2−5x+C

200

Consider the function: f(x) = x1/3

Find the linearization of f(x) at a = 8. 

f(x) = L(x) = 2 + 1/12 * (x-8)

300

What are the four statements/definitions of Rolle's theorem?

f(x) is continuous on the closed interval [a,b]

f is differentiable on the open interval (a,b)

f(a)=f(b)

Then there's another c in (a,b) such that f'(c) = 0

300
f(x) = ln(t) / 1- t

f'(x) = (1-t)(1/t) - (ln t)(-1) / (1-t)^2

300

A car's velocity is given by v(t)=3t2−4t meters per second. Find the displacement of the car from t=1 to t=3.

14 meters

300

Use the linearization you found in (a) to estimate the value of 8.1.

Remember:  L(x) = 2 + 1/12 * (x-8)

241/120

400

What is the first part of the Fundamental Theorem of Calculus about?

 It states that if a function f is continuous on [a,b], then the function g(x)=∫axf(t)dt is continuous on [a,b], differentiable on (a,b), and g′(x)=f(x).

400

g(z)= arcsin(z)+ arcsin (1/z) 

g'(z)= 1 / sqrt(1-z2) + 1 / sqrt(1-(1/z)2) * [-1/z2]

400

Find the most general antiderivative of the function g(x) = 7ex-x2/3

7ex- 3/5x5/3 + C

400

Let f(r) = x3 - 3x3 -9x + 4. Use the Closed Interval Method to find the absolute maximum and absolute minimum values of f in [0,5].

Abs max = 9

Abs min = -23

500

What does the second part of the Fundamental Theorem of Calculus allow us to do?

It provides a method for evaluating definite integrals. If F is any antiderivative of f, then ∫abf(x)dx=F(b)−F(a).

500

y = ln(arctan x)

1 / 1 + x2 // arctan(x)

500

Find the antiderivative F of f(x) = 4-3(1+x2)-1 that satisfies the condition F(1) =0.

F(x) = 4x = 3arctan(x) + 3pi - 16 // 4

500

Use logarithmic differentiation to find f'(x) where:

f(x) = 4x * e3x * (x+1)3

y'= [ ln(4)+3 + 3/x+1 ](4x * e3x * (x+1)3)