Find bounds on the real zeros of the polynomial function. \[ f(x)=-x^{4}+3 x^{3}-4 x^{2}-8 x+9 \] The lower bound is and the upper bound is
Solution
Step 1 ：First, we rewrite the polynomial function in descending order of powers of x: \(f(x) = -x^{4} + 3x^{3} - 4x^{2} - 8x + 9\).
Step 2 ：We then apply the Descartes' rule of signs to find the number of positive real roots. We count the number of sign changes in the coefficients of the polynomial: \(-1, 3, -4, -8, 9\). There are 3 sign changes, so there are 3 or 1 positive real roots.
Step 3 ：To find the number of negative real roots, we substitute \(x\) with \(-x\) in the polynomial and count the number of sign changes in the coefficients. The polynomial becomes \(-(-x)^{4} + 3(-x)^{3} - 4(-x)^{2} - 8(-x) + 9 = -x^{4} - 3x^{3} - 4x^{2} + 8x + 9\). There are 2 sign changes, so there are 2 or 0 negative real roots.
Step 4 ：Next, we use the upper and lower bounds theorem to find the bounds on the real zeros of the polynomial. The lower bound is found by substituting \(x\) with \(-x\) in the polynomial and finding the sign of the result. If the result is positive for all \(x\), then \(-x\) is a lower bound. If the result is negative for all \(x\), then \(-x\) is an upper bound. Substituting \(x\) with \(-x\) in the polynomial, we get \(-x^{4} - 3x^{3} - 4x^{2} + 8x + 9\). The result is negative for all \(x\), so \(-x\) is an upper bound.
Step 5 ：The upper bound is found by substituting \(x\) with \(x\) in the polynomial and finding the sign of the result. If the result is positive for all \(x\), then \(x\) is an upper bound. If the result is negative for all \(x\), then \(x\) is a lower bound. Substituting \(x\) with \(x\) in the polynomial, we get \(-x^{4} + 3x^{3} - 4x^{2} - 8x + 9\). The result is positive for all \(x\), so \(x\) is a lower bound.
Step 6 ：Finally, we conclude that the lower bound on the real zeros of the polynomial is \(\boxed{-\infty}\) and the upper bound is \(\boxed{\infty}\).
