Question

Which of the following is a point at which the tangent to y = -xe^(2x) is horizontal?
The answer is (-0.5,0.18) where did the 0.18 come from?

Answer 1

when u draw the graph u get y value 0.18

Answer 2

When you take the derivative of y = -xe^(2x)
\(y'=-2xe^{2x}-e^{2x}\)
you set it equal to zero
\(y'=-2xe^{2x}-e^{2x}=\)\(-e^{2x}(2x+1)\)
x= -0.5
Now, we have the x value of the tangent line, plug that back to the original equation

Answer 3

y = -xe^(2x)
plug (-0.5)
y=-(-0.5)e^(2(-0.5))
y=0.18...

Answer 4

Oh! thank you so much, both of you

Answer 5

try wolfram alpha to see the graph so u know for sure

Answer 6

You're welcome

Answer 7

you will see something like this , it really helps when u can see it !