Question

At what point on the curve y = 9 + 2ex − 5x is the tangent line parallel to the line 5x − y = 8?

Answer 1

Answer:

$(\ln(5), 19-5\ln(5))$

Step-by-step explanation:

The line can be represented as y=5x-8, so the slope of the line is 5.

The derivative of the curve is:

$\frac{d}{dx} [9+2e^x-5x]\\=\frac{d}{dx} [9]+\frac{d}{dx} [2e^x]-\frac{d}{dx} [5x]\\=2e^x-5$

We need to set that equal to 5 to find the x coordinate of our desired point

$2e^x-5=5\\2e^x=10\\e^x=5\\x=\ln(5)$

Now, just plug in x in the curve to solve for y:

$9+2e^{\ln(5) } -5\ln(5) = y\\9+10-5\ln(5)=y\\19-5\ln(5)=y$