Question 1192427
<br>
Note it is not good form to use the "√" symbol when writing your function -- we have to guess how much of the following expression is under the radical.  Use "sqrt" with parentheses around the radicand.<br>
{{{y=sqrt(4x-7)-(2/3)(x-4)=(4x-7)^(1/2)-(2/3)(x-4)}}}<br>
{{{dy/dx=(1/2)(4)(4x-7)^(-1/2)-2/3=2/sqrt(4x-7)-2/3}}}<br>
The maximum or minimum point(s) are where the derivative is zero.<br>
{{{2/sqrt(4x-7)-2/3=0}}}
{{{2/sqrt(4x-7)=2/3}}}
{{{sqrt(4x-7)=3}}}
{{{4x-7=9}}}
{{{4x=16}}}
{{{x=4}}}<br>
There is a single maximum or minimum, at x=4.<br>
{{{dy/dx=2(4x-7)^(-1/2)-2/3}}}
{{{d^2y/dx^2=(-1/2)(2)(4)(4x-7)^(-3/2) = -4/((4x-7)(sqrt(4x-7)))}}}<br>
The second derivative is always negative, so the graph has a local maximum at x=4.<br>
{{{f(4)=sqrt(4x-7)-(2/3)(x-4)=sqrt(16-7)-(2/3)(0)=sqrt(9)=3}}}<br>
ANSWER: The maximum point on the curve is (4,3)<br>
A graph, showing (not very well) the maximum point at (4,3)<br>
{{{graph(400,400,-1,7,-1,5,sqrt(4x-7)-(2/3)(x-4))}}}<br>