Answer:
b. x=2,3,4
c.
x=2 is a local minimum
x=3 is a local maximum
x=4 is a local minimum
d.
[tex]x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}[/tex]
Step-by-step explanation:
a. You have the function
[tex]f(x)=(x-2)^2(x-4)^2[/tex]
to find the intervals you have to take into account the roots of f(x), in this case you have the roots
[tex]x_{1,2}=2\\x_{3,4}=4[/tex]
- for x < 2, for example x=1:
[tex]f(1)=(1-2)^2(1-4)^2>0[/tex]
f(x) decreases from infinity to zero
- for 2 < x < 4, for example x=3:
[tex]f(3)=(3-2)^2(3-4)^2>0[/tex]
between x=2 and x=3 f(x) increases. But between x=3 and x=4 f(x) decreases because f(4)=0.
- for x > 4:
f(5)>0
f(x) increases for x > 4.
b. You have to compute the derivative to find local minima and maxima
[tex]f'(x)=2(x-2)(x-4)^2+2(x-2)^2(x-4)=2(x-2)(x-4)[(x-4)+(x-2)]\\f'(x)=2(x-2)(x-4)(2x-6)[/tex]
and by taking f'(x)=0:
[tex]f'(x)=0\\f'(x)=(x-2)(x-4)(x-3)=0\\\\x_1=2\\x_2=4\\x_3=3[/tex]
Local minima and maxima are found by evaluating the roots of f'(x) in the second derivative f''(x)
[tex]f''(x)=2[(x-4)(2x-6)+(x-2)(2x-6)+(x-2)(x-4)(2)]\\f''(x)=2[2x^2-14x+24+2x^2-10x+12+2x^2-12x+16]\\f''(x)=12x^2-72x+104[/tex]
Hence
[tex]f''(2)=12(2)^2-72(2)+104=8>0\\f''(3)=12(3)^2-72(3)+104=-4<0\\f''(4)=12(4)^2-72(4)+104=8>0[/tex]
x=2 is a local minimum
x=3 is a local maximum
x=4 is a local minimum
c.
to find the concavity you have to find the inflection points
[tex]f''(x)=0\\f''(x)=12x^2-72x+104=0\\3x^2-18x+26=0\\x=\frac{18+-\sqrt{324-4(3)(26)}}{6}\\x_1=3 + \frac{\sqrt{12}}{6}\\x_2=3 - \frac{\sqrt{12}}{6}\\[/tex]
Hence:
for -infinity < x < x1 --> concave up
for x1 < x < x2 --> concave down
for x2 < x < infinity --> concave up
d.
[tex]x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}[/tex]
