The graph of f(x) passes through (0,3). The slope of f at any point P is 5 times the y coordinate of P Find f(1)

Question

Looking for the derivative of f at 1.  

Answer 1

Amanda,

 

I almost read the question like Michael because you commented "the derivative of f at 1." What you are looking is to evaluate the original function (the antiderivative) at 1.

 

From the given:

dy  =  5y

dx

Integral dy = Integral 5 dx    move the y to dy and separate dx, then integrate
              y

 

e^ln(y) = e^5x+C

 

y = e^5x e^C      let e^C be some constant A

 

y = Ae^5x

 

3 = A e⁵⁽⁰⁾ => A=3      plug in the given coordinate (0,3)

 

 

So, f(x) = 3e^5x   and thus, f(1) = 3e⁵, which is just a number you can plug into the graphing calculator.

 

Hope this makes sense,

Marina

 

 

Answer 2

We know the equation of a line in slope-intercept form is      y=mx+b , m is slope and b is y-intercept.

 

The point (0, 3) of one of the points P.  The y-coordinate of this point is 3.  So the slope must be 15.  Therefore, the equation of the line that you have so far is

 

y = 15x + b

 

 

Plugging in the known coordinates into the formula to find b,

 

3 = 15(0) + b

 

3 = b

 

 

The equation of the line is             f(x) = 15x + 3

 

 

To find derivative of f(x) at x=1, find the derivative of f(x).  Then evaluate the derivative at x=1.

 

d/dx f(x) =  15

 

The derivative is a constant function.  So at any x value, the derivative of f(x) is 15.