Using the Intermediate Value Theorem and a calculator, find an interval of length 0.01 that contains a root of e^x = 2 - x, rounding interval endpoints off to the nearest hundredth.
The function "g(x)=e^x" is strictly increasing. The function "h(x)=2-x" is strictlydecreasing.
Then if the root of "e^x=2-x" exists, it will be the only root.
The function"f(x)=e^x-(2-x)" is continuous on "(-\\infin, \\infin)."
"f(0.5)=e^{0.5}-(2-0.5)=e^{0.5}-1.5=\\sqrt{e}-1.5>0"
"f(0.2)=e^{0.2}-(2-0.2)=e^{0.2}-1.8\\approx-0.5789<0"
"f(0.4)=e^{0.4}-(2-0.4)=e^{0.4}-1.6\\approx-0.108<0"
By the Intermediate Value Theorem exists a number "c\\in (0.4,0.5)" such that "f(c)=0"
"f(0.42)=e^{0.42}-(2-0.42)=e^{0.42}-1.58\\approx-0.058<0"
"f(0.44)=e^{0.44}-(2-0.44)=e^{0.44}-1.56\\approx-0.007<0"
By the Intermediate Value Theorem exists a number "c\\in (0.44,0.45)," such that "f(c)=0."
"(0.44,0.45)"
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