What is the instantaneous rate of change of #f(x)=3x+5# at #x_0=1#?

1 Answer

For a linear function #y=mx+b#, the rate of change, or derivative, is simply the slope m of the line. Thus, in this case the rate of change is 3 for every point along the function where the function is defined (in this case, for all real numbers #x#).

One way to prove this is to imagine what happens when we change #x# by one unit. Let us choose a generic point P, defined as #(x_1, y_1)#. If we declare #x_2 = x_1 +1#, then our #y_2# is equal to #3(x_2)+1# This is in turn equal to #3(x_1 + 1)+1# which is in turn equal to #3(x_1) +1 +3#. From our initial equation, we see that this is equal to #y_1 +3#. Thus, for every increase of #x# by one unit, #y# increases 3 units.

Another is by utilizing the power rule for functions such as #f(x)=x^n#. The power rule tells us that for these functions, the derivative #d/dx f(x) = nx^(n-1)# Since a constant function #c# has a derivative with respect to x of 0, and the function #f(x)=x# is the same as #x^1#, we know that #d/dx x = 1(x^0) = 1# Then from the constant multiple rule, we know that

#d/dx (c*f(x)) = c*(d/dx (f(x)))#

Using those equations and the Sum Rule, which states that

#d/dx [f(x)+g(x)] = d/dx f(x) + d/dx g(x)#

we can apply the power rule to a function of the sort #y=mx+b#, where b, the y-intercept, is a constant.

#dy/dx = d/dx [mx +b]#

#= d/dx mx + d/dx b#

#= m*d/dx x + d/dx b#

#= m*1(x^0) + 0#

#= m#