Study Flashcards
Differentiate this problem:
y=ln(x+1 / √x-2
1/ x+1 - 1 /√x-2 • 1 /2√x-2
Do this example:
Differentiate:
f(x)=log(10) (2+sinx)
u=2+sinx and y=log(10) (u)
f’(x)=dy/dx=dy/du • du/dx
f’(x)=1/ u(ln(10)) • cosx
f’(x)=1/ (2+sinx)(ln(10)) • cosx
=cosx/ (2+sinx)(ln(10))
How do you use the squeeze theorem for lim(x^2 (cos(7πx)) = 0
x—>0
Range of cos functions is always -1 to 1, so:
-1 <= cos(7πx) <= 1.
Multiply all by f(x) to get:
-x^2 <= x^2 (cos(7πx)) <= x^2.
All of these have limit 0 as x—>0 so by squeeze theorem, g(x) (the middle function)
approaches 0.
If you get a positive and negative limit for a problem like:
lim 8x+56 / |x+7|
x—>-7
Limit DNE
Do this example:
lim 8-|x| / 8+x
x—>-8
1 because when x is positive, the answer is unsolved and therefore DNE
Why is this undefined?:
g(x)= { 2-x^2 if 1 < x <= 2
{ x-3 if x > 2
Evaluate lim g(x)
x—>2
Because it is equal to two and not equal to two at the same time.
Do these examples:
lim [[x]] lim [[x]] lim [[x]]
x—>-5+ x—>-5 x—>-5.2
lim [[x]] lim [[x]]
x—>n- x—>n+
-5 DNE -6 n-1 n
Do this example:
Find number a such that the limit exists:
lim 3x^2 + ax + a + 6 / x^2 + x -2
x—>-2
Find the denominator number with opposite sign of limit (x+2).
Replace numerator (x) value with limit value and solve only numerator to get a.
Replace a into original equation for new equation and then solve for limit.
If lim f(x) = 0 and lim g(x)
x—>3 x—>3
does lim f(x) / g(x) exist?
x—>3
Yes
When both functions’ limits equal 0, what CAN exist?
The division of the limits of these functions
If g is continuous at a and f is continuous at g(a), then
The composite function (f o g) given by (f o g)(x) = f(g(x)) is continuous at a
The composite function (f o g) given by (f o g)(x) = f(g(x)) is continuous at a when
g is continuous at a and f is continuous at g(a)
Which types of functions are continuous at every number in their domain?
Polynomials, rationals, roots, trigs, inverse trigs, exponentials, logarithmic
Polynomials, rationals, roots, trigs, inverse trigs, exponentials, logarithmic
Functions continuous at every number in their domain
Why is the discontinuity at x=1 of
f(x)={ x+8 if x<0
{ e^x if 0 <= x <= 1
{ 9-x if x > 1
only continuous from the left?
Because there is a jump when x<0 and 0<=x<=1 in which the only actual defined points are at the second function (closed circle)
Do this example:
For what value of the constant c is the function f continuous on
(-∞, ∞)?
f(x)={ cx^2 + 2x if x< 5
{ x^3 - cx if x>= 5
Plug in 5 for x for both equation. Set these derived equations equal to each other and solve for c to get 115/30
All types
CHAR, INT, SMALLINT, VARCHAR, TINYINT, MEDIUMINT, BIGINT, DECIMAL, DATE
Statement to add constraint to column
ALTER TABLE table_name MODIFY COLUMN column_name VARCHAR(20) NOT NULL;
ALTER TABLE table_name MODIFY COLUMN column_name VARCHAR(20) NOT NULL;
Statement to add constraint to column
Make an attribute not negative
UNSIGNED constraint
Statement to return a number to 2 decimal places
SELECT TRUNCATE(num1/num2, 3);
to return to “3” decimal places
SELECT TRUNCATE(num1/num2, 3);
Returns number to 3 decimal places
Return remainder of number
SELECT attribute/number MOD 60;
Will return remainder of division by 60
SELECT attribute/number MOD 60;
Will return remainder of division by 60
