At what point on the curve #y=1+2e^x-3x# is the tangent line parallel to the line #3x-y=5#?

Question

Scarlett Allison · Accepted Answer

#(ln3,7-3ln3)# #3x-y=5# #y=3x+5# Here we have slope #m=3# since it is in the form #y=mx+b#. Since the lines are parallel, they have the same slope. Find the derivative to find the point wih tangent line of slope #3#: #y=1+2e^x-3x# #y'=0+2e^x-3# #3=2e^x-3# #3=e^x# #x=ln3# Find the #y#-coordinate. #y=1+2e^x-3x# #y=1+2e^(ln3)-3ln3# #y=7-3ln3# Thus it is parallel to the tangent line at the point #(ln3,7-3ln3)#.

Christopher Agresta · Answer

undefined To find the point on the curve where the tangent line is parallel to the line 3x-y=5, we need to determine the slope of the line 3x-y=5. The slope of this line is 3.

Next, we find the derivative of the curve y=1+2e^x-3x. The derivative is dy/dx = 2e^x - 3.

To find the x-coordinate of the point where the tangent line is parallel, we set the derivative equal to the slope of the line: 2e^x - 3 = 3.

Solving this equation, we get e^x = 3/2.

Taking the natural logarithm of both sides, we have x = ln(3/2).

Substituting this x-value back into the original equation y=1+2e^x-3x, we can find the y-coordinate of the point.

Therefore, the point on the curve where the tangent line is parallel to the line 3x-y=5 is (ln(3/2), 1+2e^(ln(3/2))-3ln(3/2)).