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Exercise 1272Calculus › Single-variable calculus › Differentiation › Mean Value TheoremL1Added 4 Nov 2026

Small-increment estimate

#MVT#small increments#estimate

Statement

Estimate 100.5\sqrt{100.5} using the MVT.

Hint

Apply the MVT to f(x)=xf(x)=\sqrt x on [100,100.5][100,100.5] with f′(c)≈12100f'(c)\approx\frac{1}{2\sqrt{100}}.

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Solution

MVT. 100.5−100=0.52c\sqrt{100.5}-\sqrt{100}=\frac{0.5}{2\sqrt c} for some c∈]100,100.5[c\in]100,100.5[, so 12100.5<100.5−100.5<12100\frac{1}{2\sqrt{100.5}}<\frac{\sqrt{100.5}-10}{0.5}<\frac{1}{2\sqrt{100}}.

Estimate. 100.5≈10+0.520=10.025\sqrt{100.5}\approx10+\frac{0.5}{20}=10.025.

Conclusion. 100.5≈10.025\sqrt{100.5}\approx10.025 (error <10−4<10^{-4}).

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