# Given the function `f(x)=(x-3)^2 (x-7)^2` , 1. The relatative maximum Value of the function is? 2. The relatative minimum Value of the function is?

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`f(x)=(x-3)^2(x-7)^2`

To determine the relative maximum and minimum of the function, take the derivative of f(x). To do so, apply the product rule which is `(u*v)' = v*u'+u*v'` .

So let,

` u= (x-3)^2` and `v=(x-7)^2`

Then, take the derivative of u and v to get u' and v'. Apply the power rule which is `(u^n)'=n*u^(n-1)*u'` .

`u'=2(x-3)^(2-1)*1=2(x-3)`

`v'=2*(x-7)^(2-1)*1=2(x-7)`

And plug-in u, v, u' and v' to the product rule of derivative.

`f'(x)=(x-7)^2*2(x-3) + (x-3)^2*2(x-7)`

Factor out the GCF to simplify f'(x).

`f'(x)=2(x-3)(x-7)[(x-7)+(x-3)]`

`f'(x)=2(x-3)(x-7)(2x-10)`

`f'(x)=2(x-3)(x-7)*2(x-5)`

`f'(x)=4(x-3)(x-7)(x-5)`

Then, set f'(x) equal to zero.

`0=4(x-3)(x-7)(x-5)`

Divide both sides by 4 to simplify the equation.

`0/4=(4(x-3)(x-7)(x-5))/4`

`0=(x-3)(x-7)(x-5)`

Then, set each factor equal to zero and solve for x.

`x-3=0 ` `x-5=0` `x-7=0`

`x=3` `x=5 ` `x=7`

These are the critical numbers of f(x).

To solve for y, plug-in the values of x to the given function.

`y=f(3)=(3-3)^2(3-7)^2=0*(-4)^2=0`

`y=f(5)=(5-3)^2(5-7)^2=2^2*(-2)^2=4*4=16`

`y=f(7)=(7-3)^2(3-3)^2=4^2*0=0`

Hence, the critical points are (3, 0), (5,16) and (7,0).

Since (5,16) has the higher value of y, it is the maximum point. And (3,0) and (7,0) are minimum points.

To check if these points are the relative maximum and minimum values of the f(x), refer to the graph of the function.

Notice that both ends of the function continuously go up, So there are other points that are higher than the critical point (5,16). Hence, (5,16) is the relative maximum point of the function.

Also,in the graph, there are no other points that is lower than (3,0) and (7,0). So, these two points are considered as the absolute minimum of f(x), not relative minimum.

**Hence,**

**(1) The relative maximum value of the function occurs at x=5,**

**(2) and the function has no relative minimum.**