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.
![x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}](https://tex.z-dn.net/?f=x_1%3D3%2B%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D%5C%5Cx_2%3D3-%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D)
Step-by-step explanation:
a. You have the function
![f(x)=(x-2)^2(x-4)^2](https://tex.z-dn.net/?f=f%28x%29%3D%28x-2%29%5E2%28x-4%29%5E2)
to find the intervals you have to take into account the roots of f(x), in this case you have the roots
![x_{1,2}=2\\x_{3,4}=4](https://tex.z-dn.net/?f=x_%7B1%2C2%7D%3D2%5C%5Cx_%7B3%2C4%7D%3D4)
- for x < 2, for example x=1:
![f(1)=(1-2)^2(1-4)^2>0](https://tex.z-dn.net/?f=f%281%29%3D%281-2%29%5E2%281-4%29%5E2%3E0)
f(x) decreases from infinity to zero
- for 2 < x < 4, for example x=3:
![f(3)=(3-2)^2(3-4)^2>0](https://tex.z-dn.net/?f=f%283%29%3D%283-2%29%5E2%283-4%29%5E2%3E0)
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
![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)](https://tex.z-dn.net/?f=f%27%28x%29%3D2%28x-2%29%28x-4%29%5E2%2B2%28x-2%29%5E2%28x-4%29%3D2%28x-2%29%28x-4%29%5B%28x-4%29%2B%28x-2%29%5D%5C%5Cf%27%28x%29%3D2%28x-2%29%28x-4%29%282x-6%29)
and by taking f'(x)=0:
![f'(x)=0\\f'(x)=(x-2)(x-4)(x-3)=0\\\\x_1=2\\x_2=4\\x_3=3](https://tex.z-dn.net/?f=f%27%28x%29%3D0%5C%5Cf%27%28x%29%3D%28x-2%29%28x-4%29%28x-3%29%3D0%5C%5C%5C%5Cx_1%3D2%5C%5Cx_2%3D4%5C%5Cx_3%3D3)
Local minima and maxima are found by evaluating the roots of f'(x) in the second derivative f''(x)
![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](https://tex.z-dn.net/?f=f%27%27%28x%29%3D2%5B%28x-4%29%282x-6%29%2B%28x-2%29%282x-6%29%2B%28x-2%29%28x-4%29%282%29%5D%5C%5Cf%27%27%28x%29%3D2%5B2x%5E2-14x%2B24%2B2x%5E2-10x%2B12%2B2x%5E2-12x%2B16%5D%5C%5Cf%27%27%28x%29%3D12x%5E2-72x%2B104)
Hence
![f''(2)=12(2)^2-72(2)+104=8>0\\f''(3)=12(3)^2-72(3)+104=-40](https://tex.z-dn.net/?f=f%27%27%282%29%3D12%282%29%5E2-72%282%29%2B104%3D8%3E0%5C%5Cf%27%27%283%29%3D12%283%29%5E2-72%283%29%2B104%3D-4%3C0%5C%5Cf%27%27%284%29%3D12%284%29%5E2-72%284%29%2B104%3D8%3E0)
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
![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}\\](https://tex.z-dn.net/?f=f%27%27%28x%29%3D0%5C%5Cf%27%27%28x%29%3D12x%5E2-72x%2B104%3D0%5C%5C3x%5E2-18x%2B26%3D0%5C%5Cx%3D%5Cfrac%7B18%2B-%5Csqrt%7B324-4%283%29%2826%29%7D%7D%7B6%7D%5C%5Cx_1%3D3%20%2B%20%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D%5C%5Cx_2%3D3%20-%20%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D%5C%5C)
Hence:
for -infinity < x < x1 --> concave up
for x1 < x < x2 --> concave down
for x2 < x < infinity --> concave up
d.
![x_1=3+\frac{\sqrt{12}}{6}\\x_2=3-\frac{\sqrt{12}}{6}](https://tex.z-dn.net/?f=x_1%3D3%2B%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D%5C%5Cx_2%3D3-%5Cfrac%7B%5Csqrt%7B12%7D%7D%7B6%7D)