Question

let ( f ) be the function given by ( f(x) = e^{-x} + cos x - 1 ). 1. what is the value of ( f(2) )?
options:
a (-1.281)
b (-1.140)
c (-1.045)
d (-1)

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = e^{-x} + \cos x - 1 \). Using derivative rules:

• Derivative of \( e^{-x} \) is \( -e^{-x} \) (chain rule: derivative of \( -x \) is \( -1 \), so \( \frac{d}{dx}e^{-x} = e^{-x} \cdot (-1) = -e^{-x} \))
• Derivative of \( \cos x \) is \( -\sin x \)
• Derivative of a constant (\( -1 \)) is \( 0 \)

So, \( f'(x) = -e^{-x} - \sin x \)

Step2: Evaluate \( f'(2) \)

Substitute \( x = 2 \) into \( f'(x) \):

\( f'(2) = -e^{-2} - \sin(2) \)

Calculate \( e^{-2} \approx \frac{1}{e^{2}} \approx \frac{1}{7.389} \approx 0.1353 \)

\( \sin(2) \approx 0.9093 \) (since \( 2 \) radians is in the second quadrant, sine is positive)

Now, \( -e^{-2} \approx -0.1353 \) and \( -\sin(2) \approx -0.9093 \)

Add them together: \( -0.1353 - 0.9093 \approx -1.0446 \approx -1.045 \)

Answer:

\( \boldsymbol{-1.045} \) (corresponding to option C)