Function problem 4

If $f(x) = x^3 +ax^2 + bx + 5sin^{2}(x)$ is increasing everywhere, then we require $f'(x) > 0$ for all $x \in \mathbb{R}.$ Taking the derivative gives:

$f'(x) = 3x^2 + 2ax + b + 5[2sin(x)cos(x)] = 3x^2 + 2ax + b + 5sin(2x) > 0$ ;

or $3x^2 + 2ax + b > -5sin(2x)$ .

Since the maximum value of $-5sin(2x)$ equals 5 for all $x \in \mathbb{R}$ we require:

$3x^2 + 2ax + b > 5$ ;

or $3x^2 + 2ax + (b-5) > 0$ ;

or $x = \frac{-2a \pm \sqrt{4a^2 - 4(3)(b-5)}}{6};$

or $x =\frac{-a \pm \sqrt{a^2 - 3b + 15}}{3}$ .

If $3x^2 + 2ax + (b-5) > 0$ is true for all real $x$ , then we require the discriminant to be negative: $a^2 - 3b + 15 < 0 \Rightarrow \boxed{a^2 - 3b < -15}.$