Let F And G Be Defined By The Table

Let f and g be defined by the table?

√(f(-3) - f(-2)) - [g(0)]² + f(-4)/g(0) ∙ g(-3) =
= √(2 - (-7)) - (-2)² + 2/(-2) ∙ 9 =
= √9 - 4 + (-1) ∙ 9 =
= 3 - 4 - 9 = -10

unless the last multiplication g(0)*g(-3) is all at the denominator of the last fraction
if so the formula is
√(f(-3) - f(-2)) - [g(0)]² + f(-4)/[g(0) ∙ g(-3)] =
= 3 - 4 + 2/[-2 ∙ 9] =
= 3 - 4 - 1/9 = -10/9

The function H is defined by the following rule.?

h(x)=4x+4
h(4)=4(4)+4=20 h(-4)=4(-4)+4= -12
h(3)=4(3)+4=16 h(-3)=4(-3)+4= -8
h(2)=4(2)+4=12

just substitute the values from the equation

Let G={x in R | x>0 and x not equal to1} Define the operation * on G by a*b=a^lnb for all a, b in G. Prove t?

note first that a^lnb = e^[(lna)(lnb)] > 0

if a and b are in G, then lna and lnb are not zero since a and b are not 1, so (lna)(lnb) is not 0, and e^[(lna)(lnb)] >0 is different than 1. So a*b is in G, and there is closure.

for a and b in G, e^[(lna)(lnb)] = e^[(lnb)(lna)], so a*b = b*a, and * is commutative.

If b=e, then lnb=1, and a*b=a^1 = a for all a. Also, e*a = e^lna = a.
So e will be our identity element.

Now x*y = e^[lnx lny], so ln(x*y) = lnx lny, and so
a*(b*c) = e^[lna ln(b*c)] = e^[lna lnb lnc]
Also, (a*b)*c = e^[ln(a*b) lnc] = e^[lna lnb lnc]
So a*(b*c) = (a*b)*c, and * is associative.

Since a is in G, lna is not 0 because a is not 1. Consider then b=e^(1/lna) > 0. b can't be 1 since 1/lna can't be 0, so b is in G.
We have lnb = 1/lna, so:
a*b = e^[lna lnb] = e^[(lna) (1/lna)] = e^1 = e
Since * is commutative, b*a = a*b =e, and e^(1/lna) is the inverse element of a in G.

So (G,*) is an abelian group.

The function h is defined by the following rule.?

Okay, so this might look complicated because of function notation (the way it's written out has h(x) but it's really actually very simple.

Your table is saying, when the value of x is -3 (first one on the table) what is the value that the function will produce.

To solve this, just sub in the value for x in the x in your equation
Ex. When x is -3
h(x)=2x+3
h(x)=2(-3) + 3
h(x)= -6 + 3
h(x)= -3

What i'm doing is subbing in the x value in the table, for the x value in the function. Another example I will do is 2.

When x = 2
h(x)=2x+3
h(x)=2(2) +3
h(x)= 4 + 3
h(x) = 7

Now that wasn't so hard was it :)
Hope that helps

Let f and g be differentiable functions?

Let f and g be differentiable functions and let the values f, g, and the derivatives f' and g' at x = 1 and x = 2 be given by the table below:

x f(x) g(x) f'(x) g'(x)
1 3 2 5 4
2 2 pi 6 7

Determine the value of each of the following:
a. the derivative of f + g at x=2
b. the derivative of fg at x=2
c. the derivative of f/g at x=2
d. h'(1) where h(x) = f(g(x))

I need help on an algebra 2 problem?

The linear function defined by the table is y = 2x + 7 since its y intercept is given to be 7, and its slope is (7 - -3)/(0 - -5) = 10/5 = 2
So f(x) = 2x + 7; to do a horizontal stretch of 2.3, divide the x coefficient by 2.3
g(x) = 20/23 x + 7

That's my opinion anyway, see what you think

Let f, g, and their derivatives...?

1.
f(1) = 3, f(2) = 0, f(3) = 1, f(4) = 2
g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 1
f'(1) = -1, f'(2) = -1, f'(3) = 0, f'(4) = 2
g'(1) = -2, g'(2) = 2, g'(3) = 1, g'(4) = -1

h(1) = f(g(1)) = f(2) = 0
h(2) = f(g(2)) = f(3) = 1
h(3) = f(g(3)) = f(4) = 2
h(4) = f(g(4)) = f(1) = 3

By the Chain Rule, h'(c) = f'(g(c)) * g'(c), therefore:

h'(1) = f'(g(1)) * g'(1) = f'(2) * g'(1) = (-1)(-2) = 2
h'(2) = f'(g(2)) * g'(2) = f'(3) * g'(2) = (0)(2) = 0
h'(3) = f'(g(3)) * g'(3) = f'(4) * g'(3) = (2)(1) = 2
h'(4) = f'(g(4)) * g'(4) = f'(1) * g'(4) = (-1)(-1) = 1

Answer: h(3) = h'(3) = 2

2.
y = sqrt(8 - x^2)
Let u = 8 - x^2, so du/dx = -2x
So y = sqrt(u) = u^0.5, so dy/du = 0.5u^(-0.5) = 1 / (2 * sqrt(8 - x^2))
By the Chain Rule:
dy/dx = dy/du * du/dx
dy/dx = (1 / (2 * sqrt(8 - x^2))) * (-2x)
dy/dx = -x / sqrt(8 - x^2)
If x = 2 we have:
dy/dx = -2 / sqrt(8 - 2^2)
dy/dx = -2 / sqrt(8 - 4)
dy/dx = -2 / sqrt(4)
dy/dx = -2 / 2
dy/dx = -1
So the slope of the tangent line is -1.
Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of -1, which is 1.

The function g(x) is a continuous quadratic function defined for all real #s, with some of its values given by the table below.?

The quadratic function f(x) is represented by the parabola below [see attached]


Select all the statements that are TRUE.

a) the function f(x) has a maximum value at x=4.5

b) the function g(x) has a minimum value of 0

c) the maximum value of g(x) is twice the maximum value of f(x)

d) neither function has a minimum value

How do you do this Algebra 2 problem. please help!?

g(x) = f(x) + 1.5

so if f(x) was equal to a hypothetical x^2

G(x) would be x^2 + 1.5


Remember, for vertical translation, all you do is add or subtract the desired number of units to the overall function