Question 1107335
1) {{{f(x)=sqrt(25-x^2)}}} is define only for {{{-5<=x<=5}}}
{{{f(4)=sqrt(25-4^2)=sqrt(25-16)=sqrt(9)=3}}}
 
As a calculus problem:
The derivative of {{{f(x)=sqrt(25-x^2)=(25-x^2)^"1/2"}}}
{{{df/dx=(1/2)(25-x^2)^"-1/2"(-2x)=-x/sqrt(25-x^2)}}}
For {{{x=4}}} , {{{df/dx}}}{{{(4)=-4/sqrt(25-16)=-4/sqrt(9)=-4/3}}}
is the slope of the tangent.
 
Otherwise,
{{{y=sqrt(25-x^2)}}} is the {{{y>=0}}} half of the circle with equation
{{{y^2=25-x^2}}} <--> {{{x^2+y^2=5^2}}}
The tangent at the point with {{{system(x=4,y=3)}}}
is perpendicular to the radius at that point.
The equation of the line containing that radius is {{{y=(3/4)x}}} .
The slope of a line perpendicular to that radius is {{{-1/"3/4"=-4/3}}}
 
Either way, the equation of the tangent
at the point with {{{system(x=4,y=3)}}} , with slope {{{-4/3}}} is
{{{y-3=(-4/3)(x-4)}}}
{{{y-3=(-4/3)x+16/3}}}
{{{y=(-4/3)x+16/3+3}}}
{{{highlight(y=(-4/3)x+25/3)}}}
{{{drawing(300,300,-6,6,-1,11,grid(0),
blue(arc(0,0,10,10,180,360)),green(line(-3,37/3,7,-1))
)}}}
 
2) {{{f(x)=x^2+sqrt(x)=x^2+x^"1/2"}}} is defined for {{{x>=0}}} .
{{{f(1)-1^2+sqrt(1)=1+1=2}}}
{{{df/dx=2x+(1/2)x^"-1/2"=2x+1/2sqrt(x)}}}
The slope of the tangent at {{{x=1}}} is
{{{df/dx}}}{{{(1)=2*1+1/2sqrt(1)=2+1/2=5/2}}}
The equation of the tangent line
passing through the point with {{{system(x=1,y=2)}}} ,
with slope {{{5/2}}} is
{{{y-2=(5/2)(x-1)}}}
{{{y-2=(5/2)x-5/2}}}
{{{y=(5/2)x-5/2+2}}}
{{{highlight(y=(5/2)x-1/2)}}}
{{{graph(300,300,-1,4,-1,9,-2,(5/2)x-1/2,x^2+sqrt(x))}}}