Use continuity and the intermediate value theorem to solve problems. If f(x)=x3−6x+8, show that there are values c for which f(c) equals a)π, b) −3, and c) 6,000,000.

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Use continuity and the intermediate value theorem to solve problems.  If f(x)=x3−6x+8, show that there are values c for which f(c) equals a)π, b) −3, and c) 6,000,000.

Momp1989 · Accepted Answer

Given f(x)=x3−6x+8 Note that intermediate value theorem states that if f is continuous on a domain containing an interval [a,b] and let be s be number with f(a) Since f(x)=x3−6x is polynomial, it is continuous everywhere on the real line. That is, f(x)=x3−6x+8 is continuous on the interval (−∞,∞) Also, the range of f(x)=x63x+8 is (−∞,∞). a) Note that π lies in the interval (−∞,∞). Simce f is continuous everywhere, f satisfies the intermediate value theorem. Therefore, the exist a c such that f(x)=π b) Note that −3 lies in the interval (−∞,∞). Since f is continuous everywhere, f satisfies the intermediate value theorem. Therefore, there exist a c such that f(c)=−3 c) Note that 6,000,000 lies in the interval (−∞,∞). Since f is continuous everywhere, f satisfies the intermediate value theorem. Therefore, there a c such that f(c)=6,000,000

Ched1950 · Accepted Answer

Given, f(x)=x3−6x+8 a) Since f(x) is a polynomial in x, f(0)=0+6×0+8 f(0)=8 Puting x=1 f(1)=1−6+8 =9−6 =3 f(x) is continuous on [0,1] f(1)&amp;lt;π&amp;lt;f(0) Hence exists 0 f(c)=π b) puting x=0 f(0)=8 puting x=-4 f(−4)=(−4)3−6×(−4)+8 =−32 f(x)=x3−6x+8 is continuous on [-4,0] ∴f(−4)&amp;lt;−3&amp;lt;f(0) −4&amp;lt;c&amp;lt;0, c = real number f(c)=−3 c) putting x=0 f(0)=8 puting x=1000 f(1000)=(1000)3−6(1000)+8 =999,994,008 f(x) is continuous on [0,1000] ∴f(0)&amp;lt;6,000,000&amp;lt;f(1000) c is rea not exist as 0&amp;lt;c&amp;lt;1000 f(c)=6,000,000

Use continuity and the intermediate value theorem to solve problems. If f

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