Question

attempt 1: 10 attempts remaining. find the critical values of the given function. list all values separated by commas. ( f(x) = -2e^{-x} ) ( x = ) submit answer

Explanation:

Step1: Recall the definition of critical values

Critical values of a function \( f(x) \) are the values of \( x \) where \( f'(x) = 0 \) or \( f'(x) \) does not exist. First, we need to find the derivative of \( f(x)=-2e^{-x} \).

Step2: Find the derivative of \( f(x) \)

Using the chain rule, the derivative of \( e^{u} \) with respect to \( x \) is \( e^{u}\cdot u' \). For \( f(x)=-2e^{-x} \), let \( u = -x \), then \( u'=-1 \). So the derivative \( f'(x) \) is:
\( f'(x)=-2\cdot e^{-x}\cdot(-1) = 2e^{-x} \)

Step3: Analyze when \( f'(x) = 0 \) or does not exist

The exponential function \( e^{-x}=\frac{1}{e^{x}} \) is never zero for any real number \( x \), and it is defined for all real numbers \( x \). So we set \( f'(x) = 0 \):
\( 2e^{-x}=0 \)
But \( e^{-x}>0 \) for all real \( x \), so \( 2e^{-x}>0 \) for all real \( x \). Therefore, there are no solutions to \( f'(x) = 0 \). Also, \( f'(x) = 2e^{-x} \) is defined for all real \( x \) (since the exponential function is defined everywhere). So we check if there are any points where \( f'(x) \) is undefined, but since \( e^{-x} \) is defined for all \( x \), \( f'(x) \) is defined for all \( x \). Wait, but maybe we made a mistake? Wait, let's re - check the derivative.

Wait, the original function is \( f(x)=-2e^{-x} \). The derivative of \( e^{-x} \) with respect to \( x \) is \( -e^{-x} \), so \( f'(x)=-2\times(-e^{-x}) = 2e^{-x} \). Now, \( e^{-x}=\frac{1}{e^{x}} \), and \( e^{x}>0 \) for all real \( x \), so \( 2e^{-x}>0 \) for all real \( x \). So \( f'(x) \) is never zero and is defined everywhere. So does that mean there are no critical values?

Wait, but maybe the function is a constant? No, \( f(x)=-2e^{-x} \) is not a constant. Wait, let's think again. The critical points are where the derivative is zero or undefined. Since the derivative is \( 2e^{-x} \), which is always positive and never zero, and is defined for all real numbers, there are no critical values.

Answer:

There are no critical values (or we can say there is no solution in the set of real numbers). But if we follow the problem's requirement to list values (even if none), we can state that there are no critical values. However, maybe we made a mistake in the derivative? Wait, let's re - derive:

\( y = - 2e^{-x}\)

Using the formula \( \frac{d}{dx}(e^{ax})=ae^{ax} \), here \( a=-1 \), so \( \frac{d}{dx}(-2e^{-x})=-2\times(-1)e^{-x}=2e^{-x} \). Yes, the derivative is correct. And \( 2e^{-x}=0 \) has no solution because \( e^{-x}>0 \) for all \( x\in R \). So the function \( f(x) = - 2e^{-x}\) has no critical values.