i received an assignment and i am unsure of where to start. the question i cant solve is this: Given $\displaystyle p(x) = x^3 - x + 1$, find all values of a such that $\displaystyle p (a - 1) > p(1)$
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Originally Posted by plasticfang Given $\displaystyle p(x) = x^3 - x + 1$, find all values of a such that $\displaystyle p (a - 1) > p(1)$ First expand: $\displaystyle p\left( {a - 1} \right) = \left( {a - 1} \right)^3 - \left( {a - 1} \right) + 1 = a^3 - 3a^2 + 2a + 1$ Now solve $\displaystyle a^3 - 3a^2 + 2a + 1 > 1 = p(1)$ for a.
am i doing the right thing if i go $\displaystyle a(a^2 - 3a + 2) > 0$ $\displaystyle a(a - 1)(a - 2) > 0$
Originally Posted by plasticfang am i doing the right thing if i go $\displaystyle a(a^2 - 3a + 2) > 0$ $\displaystyle a(a - 1)(a - 2) > 0$ YES! You have it! Why don't you finish it off?
ok just got back from work. so my a values would be 0 1 and 2?
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