We need to find the maximum and minimum points on [-2,1] for the function
[tex]f(x)=2x⁵+5x⁴[/tex]
Then
[tex]f^{\prime}(x)=10x⁴+20x³[/tex]
Now to find the critical points we have that
[tex]f^{\prime}(x)=0\leftrightarrow10x⁴+20x³=0[/tex]
so
[tex]10x⁴+20x³=10x³(x+2)=0\text{ }\leftrightarrow x=0\text{ or }x=-2[/tex]
We get that we have two critical points at the interval [-2,1].
Now, to determine if they are minimum or maximum points we choose any point before and after them. We will choose -3,-1,1.
So,
[tex]f^{\prime}(-3)=10(-3)⁴+20(-3)³=10(81)-20(27)=810-540>0[/tex][tex]f^{\prime}(-1)=10(-1)⁴+20(-1)³=10-20=-10<0[/tex][tex]f^{\prime}(1)=20(1)³+10(1)⁴=30>0[/tex]
We should notice that f'(-3)>0 f'(-2)=0 and f'(-1)<0. Then
[tex]x=-2\text{ is a maximum point}[/tex]
Moreover, since f'(-1)<0 f(0)=0, f'(1)>0. Then
[tex]x=0\text{ is a minimum point }[/tex]
(a) We have that x=-2 is a maximum point and x=0 is a minimum point on the interval [-2,1]
To find the intervals where the function increase and decrease, notice that
[tex]f^{\prime}(x)>0,\text{ for }x\in(-\infty,-2).\text{ Then }f\text{ increases on }(-\infty,-2).[/tex]
since f'(x) is a continuous polynomial and f'(-2)=0 and f'(-3)>0.
Now,
[tex]f(x)\text{ decreases on }(-2,0)[/tex]
since f'(-2)=0 and here attaches a maximum, then the function needs to decrease until the nex critical point that is the point x=0.
After taking the minimum point x=0, f(x) will increase, i.e.
[tex]f(x)\text{ increases on }(0,\infty)[/tex]
(b) f(x) increases on (-infinity,-2), decreases on (-2,0) and increases once more time on (0,infinity)
[tex]f(x)=0\leftrightarrow x⁴(2x+5)=0\leftrightarrow x=0\text{ or }x=\frac{-5}{2}[/tex]
So,
