The solution is \displaystyle{x}\in{\left[-{3},{2}\right]}\cup{\left[{4},+\infty\right)} Explanation: Let \displaystyle{f{{\left({x}\right)}}}={\left({x}-{4}\right)}{\left({x}+{3}\right)}{\left({2}-{x}\right)} ...

See a solution process below: Explanation: First, subtract \displaystyle{\left({24}\right)} and add \displaystyle{\left({8}{x}\right)} to each side of the inequality to isolate the \displaystyle{x} ...

When I want to check my computation of a Fourier series, I just ask Wolfram Alpha to plot it : The plot shows that you got the curve mostly right, but it's too low under the axis. Look at the ...

For x=0 to be the coordinate of the local maximum, there must exist a neighborhood of 0 such that f(0) is the maximum. Since cosine attains 1 when x=0, a sufficiently small right neighborhood of ...

Let A=\{f\in \mathcal C([0,1],[-1,1])\;,\; \forall x, |f(x)|\leq x\} . It is not hard to prove that \sup_{f\in A}\left|\int_0^1f^2-f \right|= \max\left(\sup_{f\in A}\left[\int_0^1f^2-f\right], -\inf_{f\in A}\left[\int_0^1f^2-f\right] \right) ...

Yes your proof is correct. Excellent work reducing the question about multiplying inequalities to a more familiar one of adding inequalities. The only thing I would mention is that taking logarithms ...