Question
Prove that the function $g$ defined for all real $x$ by $g(x)=$ $e^x-1-x$ has a unique minimum value of zero when $x=0$.
Step 1
To find the minimum value of the function $g(x) = e^x - 1 - x$, we start by finding its first derivative. Using basic calculus rules, we have: \[ g'(x) = \frac{d}{dx}(e^x - 1 - x) = e^x - 1. \] Show more…
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Key Concepts
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