At what point on the curve y = 2 + 2e^x − 5x is the tangent line parallel to the line 5x − y = 9?

Question

(x,y) = ?

Please show work and answer so i can grasp a better understanding for this type of problem

Answer 1

The derivative of y gives the slope of the line tangent to that function at any point "x" so, let's find the derivative:

y = 2 + 2e^x − 5x

y' = 2e^x − 5 (using the exponential rule and the power rule)

The line 5x - y = 9 can easily by manipulated to be in slope-intercept form:

y = 5x - 9 and we can see the slope is 5. A line parallel to it will also have the slope 5 so:

2e^x − 5 = 5

2e^x = 10

e^x = 5

ln(e^x) = ln(5)

xln(e) = ln(5)

x(1) = ln(5)

x = ln(5)

ln(5) can be approximated by plugging it into your calculator as approx 1.609 but it is better to leave it as ln(5).

If x = ln(5), we need to find the y-value of the point. We can do so by plugging it into the original function:

y = 2 + 2e^x − 5x

y = 2 + 2e^(ln(5)) − 5(ln(5))

y = 2 + 5•2 − ln(5⁵)

y = 12 - ln(3125)

So the point is (ln(5), (12 - ln(3125))