f(x) passes through (0,2). the slope of f at any point is 3 times the y-coordinate. find the derivative at 1
When a derivative of a function is a multiple of the y value of the function, it suggests an exponential function.
let the function be y = a e^(kx), where a and k are constanst
dy/dx = ak e^(kx)
given ak e^(kx) = 3y = 3(a e^(kx))
then ak = 3a
k = 3
so we know y = a e^(3x)
at (0,2)
2 = a e^(0)
a = 2
then y = 2 e^(3x)
then dy/dx = 6 e^(3x)
when x=1
dy/dx = 6 e^3
dy/dx = 3 y
dy/y = 3 dx
ln y = 3 x + c
y = C e^3x
if x = 0, y = 2
so
y = 2 e^3x
when x = 1
y = 2 e^3 = 40
dy/dx = 3y = 120
To find the derivative at a point on a function, we need to start by finding the derivative function.
Given that the slope of the function at any point is 3 times the y-coordinate, we can express this relationship as:
dy/dx = 3y
We can solve this differential equation by separating the variables and integrating both sides. Let's do that.
dy/y = 3dx
Integrating both sides:
∫(1/y)dy = ∫3dx
ln|y| = 3x + C
Next, we can use the initial condition that f(x) passes through the point (0, 2), which means x = 0 and y = 2. Substituting these values into the equation, we can solve for C.
ln|2| = 3(0) + C
ln|2| = C
So the equation becomes:
ln|y| = 3x + ln|2|
Exponentiating both sides:
|y| = e^(3x + ln|2|)
Since we are working with y as the dependent variable, we can drop the absolute value:
y = ±e^(3x + ln|2|)
Now, to find the derivative at x = 1, we can substitute this value into the derivative function:
dy/dx = 3y
dy/dx = 3(±e^(3x + ln|2|))
dy/dx = ±3e^(3x + ln|2|)
Now, to find the derivative at x = 1, substitute x = 1 into the equation:
dy/dx = ±3e^(3(1) + ln|2|)
dy/dx = ±3e^(3 + ln|2|)
So the derivative at x = 1 is ±3e^(3 + ln|2|).