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

Calc12

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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.
 
Calc12 said:
A) y=ex\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

y=ex\displaystyle y'=-e^{-x}

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

At x=-1, y=e.

So, using y=mx+b:

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

b=0\displaystyle b=0

The line passes through the origin and has equation

y=ex\displaystyle y=-e\cdot x

y=2.718x\displaystyle y=-2.718x

B)y=2ex\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


y=ex\displaystyle y'=e^{-x}

The slope at x=1 is m=e1=1e\displaystyle m=e^{-1}=\frac{1}{e}

At x=1,   y=2e1=21e\displaystyle x=1, \;\ y=2-e^{-1}=2-\frac{1}{e}

Thus, 21e=1e(1)+bb=22e\displaystyle 2-\frac{1}{e}=\frac{1}{e}(1)+b\Rightarrow b=2-\frac{2}{e}

y=1ex+22e\displaystyle y=\frac{1}{e}x+2-\frac{2}{e}

Here's the graph of this last one:
 
Thanks so much for clarifying, correcting and validating my work!
Really really appreciate it.
 
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