# What are the critical values, if any, of #f(x)=x+e^(-3^x)#?

Therefore f' is positive for all values of x.

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To find the critical values of (f(x) = x + e^{-3^x}), we need to find where the derivative of (f(x)) is zero or does not exist. The derivative of (f(x)) is:

[f'(x) = 1 - 3^x \cdot e^{-3^x} \cdot \ln(3)]

Setting (f'(x)) to zero and solving for (x) gives:

[1 - 3^x \cdot e^{-3^x} \cdot \ln(3) = 0]

Since (e^{-3^x}) is always positive, the only way for the expression to be zero is when (1 - 3^x = 0). Solving this gives (x = \log_3(1)), which is not a real number. Therefore, there are no critical values for (f(x)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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