F(x) = (2/3)x+3
F'(x) = (2/3)x-2
The car was bought for $10,000 in 2000. The price in 2010 decreased to $5,000.
Find the value of a car in 2023 using an exponential model.
$2030.63
Determine composition of functions for w(x).
g(x) = x-1
f(x) = 2x
f(g(x))
Find the quadratic equation for g(x)
g(x) = -2(x+1)2+3
F(x) = e(2x-3)
F'(x) = 1 + ln(x)/3
Solve this exponential using Desmos:
F(x) = 2*(e)x
F(x) = 4
On desmos type the following:
y = 2ex
y = 4
They should cross at x = 0.693.
Using h(x) and w(x) solve c(x)
c(x) = h(x) * w(x)
2
Find the intervals where k(x) is decreasing when the domain is restricted to [-2,4].
(-2,-1.5)U(-0.5,0.5)U(1.5,2.5)U(3.5,4)
Find the inverse of the following:
h(x)
h'(x) = ln(x/4)/ln(1/2)
Solve the Exponential Using Definition of Log:
y = 23(1/3)x-5
y = 25
x = 5+ log1/3(25/23)
or
x = 5 + ln(25/23)/ln(1/3)
x = 4.924
Simplify the following j(x) = z(x) - y(x)
j(x) = x2+3x+2
Solve z(x) = 0
Not a solution
Find the inverse Function:
F(x) = log(e100x-10)
Solve using logs on both sides:
5x+y= e7x
Solve for y
y= 7x/ln(5) -x
Find f(w(y(x)))
-3(2-3w)+1
What is the domain and range of k(x)
Domain: (-infinity, infinity)
Range: [-3,5]