Find the equation of the tangent line to the curve ..

[Calc12]

A) y = e^(-x) @ x = -1

dy/dx = (-1)(1)(e^(-x))
= -e^(-x)

f'(-1) = -0.368
f(-1) = 0.368

y =mx+b
0.368 = (-0.38)(-1) + b
0=b

y=-0.368x

B) y = 2 - e^(-x) @ x=1

dy/dx = 0 +(-1)(-e^(-x))
= e^(-x)

f'(1) = -0.368
f(1) = 1.632

y=mx+b
1.632 = (-0.38)(1)+b
2 = b

y= -0.368x + 2

Is this correct? I feel funny about it.

 

[galactus]

A) \(\displaystyle y = e^{-x}\) @ x = -1

dy/dx = (-1)(1)(e^(-x))
= -e^(-x)

f'(-1) = -0.368
f(-1) = 0.368

y =mx+b
0.368 = (-0.38)(-1) + b
0=b

y=-0.368x

\(\displaystyle y'=-e^{-x}\)

Thus, the slope at x=-1 is \(\displaystyle -e^{-(-1)}=-e\)

At x=-1, y=e.

So, using y=mx+b:

\(\displaystyle e=-e\cdot (-1)+b\)

\(\displaystyle b=0\)

The line passes through the origin and has equation

\(\displaystyle y=-e\cdot x\)

\(\displaystyle y=-2.718x\)

B)\(\displaystyle y = 2 - e^{-x}\) @ x=1

dy/dx = 0 +(-1)(-e^(-x))
= e^(-x)

f'(1) = -0.368
f(1) = 1.632

y=mx+b
1.632 = (-0.38)(1)+b
2 = b

y= -0.368x + 2

Is this correct? I feel funny about it.Calc12

\(\displaystyle y'=e^{-x}\)

The slope at x=1 is \(\displaystyle m=e^{-1}=\frac{1}{e}\)

At \(\displaystyle x=1, \;\ y=2-e^{-1}=2-\frac{1}{e}\)

Thus, \(\displaystyle 2-\frac{1}{e}=\frac{1}{e}(1)+b\Rightarrow b=2-\frac{2}{e}\)

\(\displaystyle y=\frac{1}{e}x+2-\frac{2}{e}\)

Here's the graph of this last one:

 

[Calc12]

Thanks so much for clarifying, correcting and validating my work!
Really really appreciate it.