For what values of x does the graph of #f(x)=e^(3x)# and #g(x)=x^2-2# have a horizontal tangent line?

Question

Christopher Allers · Accepted Answer

The graph of #f(x) = e^(3x)# does not have a horizontal tangent line. The graph of #g(x) = x^2 -2# has a tangent line at #x=0# The two functions do not share a tangent line (which seems to be what you are asking for). Only read past this point to see where I got the tangent line value (for #g(x)#) and the non-existence of a tangent line (for #f(x)#): "horizontal tangent line" #rarr# slope #= 0# For #g(x)# (always do the easy one first)
the slope at #x# is given by the derivative of #g(x)#
#g'(x) = 2x# for a slope of #0#
#g'(x) = 2x = 0#
#rarr x = 0# For #f(x) = e^(3x)#
#f'(x) = (d (3x))/dx * e^(3x)#
#f'(x) = 3 e^(3x)# but
#e^(anynumber)# is always #>0#
so
#f(x)# does not have a horizontal tangent

Ezra Adams · Answer

undefined To find the values of x where the graph of f(x) = e^(3x) and g(x) = x^2 - 2 have a horizontal tangent line, we need to find the points where the derivative of each function is equal to zero.

The derivative of f(x) = e^(3x) is f'(x) = 3e^(3x). Setting this equal to zero, we get 3e^(3x) = 0. However, since e^(3x) is always positive, there are no values of x that make f'(x) equal to zero. Therefore, the graph of f(x) = e^(3x) does not have a horizontal tangent line.

The derivative of g(x) = x^2 - 2 is g'(x) = 2x. Setting this equal to zero, we get 2x = 0. Solving for x, we find x = 0. Therefore, the graph of g(x) = x^2 - 2 has a horizontal tangent line at x = 0.