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# (a) Prove that the equation has at least one real root.(b) Use your calculator to find an interval of length 0.01 that contains a root.$\cos x = x^3$

## (a) $f(x)=\cos x-x^{3}$ is continuous on the interval $[0,1], f(0)=1>0,$ and $f(1)=\cos 1-1 \approx-0.46<0 .$ since$1>0>-0.46,$ there is a number $c$ in (0,1) such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a rootof the equation $\cos x-x^{3}=0,$ or $\cos x=x^{3},$ in the interval (0,1)(b) $f(0.86) \approx 0.016>0$ and $f(0.87) \approx-0.014<0,$ so there is a root between 0.86 and $0.87,$ that is, in the interval(0.86,0.87)

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