Extremal value problems: Difficult Problems with Solutions

Solution:
We rewrite [tex]f(x)[/tex] as [tex]f(x)=-x^2+18x+3=-x^2+2 \times 9 \times x-81+81+3=-(x^2-2 \times 9 \times x+9^2)+84=-(x-9)^2+84[/tex]. Since [tex]-(x-9)^2 \le 0[/tex], [tex]f(x) \le 84[/tex] with equality for [tex]x=9[/tex].

Solution:
[tex]f(x)=\sqrt{2}|sinx+cosx|=\sqrt{2}|\sqrt{2}\frac{\sqrt{2}}{2}(sinx+cosx)|=2|\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx|=2|\sin\frac{\pi}{4}sinx+\cos\frac{\pi}{4}cosx|=2|cos(\frac{\pi}{4}-x)|[/tex]. We know that [tex]|cosu| \le 1[/tex], so [tex]2|cos(\frac{\pi}{4}-x)| \le 2.1=2[/tex].

Solution:
[tex]1+|x-1| \ge 1[/tex], with equality for [tex]x=1[/tex]. So [tex]|1+|x-1|| \ge |1|=1[/tex], since [tex]1[/tex] is a positive number.

Solution:
[tex]f(x)=x^2+2 \times \frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+1=(x+\frac{5}{2})^2-\frac{21}{4}[/tex]. Since [tex](x+\frac{5}{2})^2 \ge 0[/tex], the minimal value for [tex]f(x)[/tex] is [tex]-\frac{21}{4}[/tex].

Solution:
The extrema of [tex]f(x)[/tex] are reached for the values of [tex]x[/tex], for which its first derivative is zero. [tex]f'(x)=2x+8[/tex], which is zero for [tex]x=-4[/tex]. So, an extremum for [tex]f(x)[/tex] is [tex]f(-4)=(-4)^2+8 \times (-4)+15=16-32+15=-1[/tex]. Since the graph of [tex]f(x)[/tex] is a parabola with a positive leading coefficient, the extremum is a minimum.

Solution:
Since the function is a quadratic function with a negative leading coefficient, its only extremum is a maximum. It is also reached when x equals the root of the equation [tex]f'(x)=0[/tex]. We calculate [tex]f'(x)=-2x-10[/tex], which has a root [tex]x=-5[/tex].

Solution:
[tex]f(x)[/tex] is a quartic function with a positive leading coefficient, so it has a minimal value. Its first derivative is [tex]f'(x)=x^3+3x^2+4x-8=x^3-x^2+4x^2+4x-8=x^2(x-1)+4(x^2+x-2)=x^2(x-1)+(4x+8)(x-1)=(x-1)(x^2+4x+8)[/tex]. One root of the derivative is [tex]x=1[/tex]. Since [tex]x^2+4x+8=x^2+4x+4+4=(x+2)^2+4>0[/tex], it has no other roots. So the minimum is reached for [tex]x=1[/tex] and it is [tex]f(1)=\frac{41}{4}[/tex].

Solution:
[tex]f'(x)=x^3-6x^2+11x-6=x^3-x^2-5x^2+5x+6x-6=x^2(x-1)-5x(x-1)+6(x-1)=(x-1)(x^2-5x+6)=(x-1)(x-2)(x-3)[/tex]. Since [tex]f(x)[/tex] is a quartic function with a positive leading coefficient, then it reaches its minimal value in a root of [tex]f'(x)[/tex]. Since [tex]f'(x)[/tex] has three roots ([tex]x_1=1,x_2=2,x_3=3)[/tex], we substitute each of them and calculate the value of [tex]f(x)[/tex]: [tex]f(1)=\frac{23}{4}[/tex], [tex]f(2)=6[/tex] and [tex]f(3)=\frac{23}{4}[/tex]. So the function has a minimal value of [tex]\frac{23}{4}[/tex], which is reached for [tex]x=1[/tex] and [tex]x=3[/tex].

Solution:
We rewrite [tex]f(x)=cos^2x+2cosx+2(1-cos^2x)=cos^2x+2cosx+2-2cos^2x=-cos^2x+2cosx+2[/tex]. We now substitute [tex]cosx=t \in [-1;1][/tex] and get [tex]g(t)=f(x)=-t^2+2t+2, t \in [-1;1][/tex]. But [tex]g(t)=-t^2+2t+2=-t^2+2t-1+3=-(t-1)^2+3[/tex], which is maximal for [tex]t=1[/tex] and [tex]t=1[/tex] belongs in the given interval, so the maximal value for [tex]f(x)[/tex] is [tex]3[/tex].

Solution:
[tex]f(t)=log_2(t)[/tex] is a strictly increasing function for all [tex]t > 0[/tex], so its minimal value is reached in the minimal value of the argument. But [tex]|x|+2 \ge 2[/tex] (since [tex]|x|[/tex] is non-negative), therefore [tex]log_2(|x|+2) \ge log_2(2)=1[/tex]