Hello!

Note that x>0 for ln(x/4) to make sense.

To find a relative extremum we have to solve equation y'(x)=0

and to find a point of inflection — equation y''(x)=0.

Also note that [ln(x/4)]' = 1/x

(because `[ln(x/4)]' = (1/4)*(1/(x/4))` or because ln(x/4) = lnx - ln4).

Now, let's find y' and y'':

`y'(x) = 7*[x^2*ln(x/4)]' = 7*[2x*ln(x/4) +x^2/x] = 7*[2x*ln(x/4) + x]` .

`y''(x) = 7*[2x*ln(x/4) + x]' = 7*[2ln(x/4) + (2x)/x + 1] = 7*[2ln(x/4) + 3]` .

So, y'(x)=0 means `7*[2x*ln(x/4) + x] = 7x*[2ln(x/4) + 1] = 0` .

Therefore x=0 (formally not suitable) or

`2ln(x/4) = -1` , `ln(x/4) = -1/2` , `x/4 = e^(-1/2)` , `x=x_1 = 4e^(-1/2) approx 2.43` .

And y''(x)=0 means `2ln(x/4) = -3` , `x/4 = e^(-3/2)` , `x=x_2=4e^(-3/2) approx 0.89` .

Also we are asked to find `y(x_1)` and `y(x_2)` . This is also simple:

`y_1 = y(x_1)` =

`7*(x_1)^2*ln(x_1/4) = 7*(16*e^(-1))*ln(e^(-1/2)) =`

` 7*16*e^(-1)*(-1/2) = -56*e^(-1) approx -20.60.`

`y_2 = y(x_2) = 7*(x_2)^2*ln(x_2/4) = 7*(16/e^3)*ln(e^(-3/2)) =`

` 7*(16/e^3)*(-3/2) = -168/e^3 approx -8.36.`

The answer:

relative extremum **(2.43, -20.60)**

point of inflection **(0.89, -8.36)**

Also please look at the graph here: https://www.desmos.com/calculator/kyndorhaiv

(also I must note that there is a limit of y(x) as x->+0 and it is 0. And y may be defined at 0 as y(0)=0 and y will be continuous and differentiable at 0, and x=0 will be one more extremum)