Discussion
If \(f'(x) = e^{x}\) and \(f(0) = 5\), then from Mean Value Theorem, the value of \(f(1)\) lies between
Answer: D
\(f(1)\) lies between 6 and (5 + e).
By the Mean Value Theorem there is some c with 0 < c < 1 such that
\[ \begin{aligned} f(1) - f(0) &= f'(c)(1 - 0) = e^{c} \\ 1 &\lt e^{c} \lt e \\ 6 &\lt f(1) \lt 5 + e \end{aligned} \]
Indeed \(f(x) = e^{x} + 4\) fits both conditions and gives \(f(1) = 4 + e \approx 6.72\), inside that interval. The other options forget to add \(f(0) = 5\), or add the wrong constant, so their bounds do not contain the value.
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