تطبيقات على التفاضل : حلول الاسئلة

We know that the tangent is the first derivative  of \(f(x)\) at \(x=0\) then ,
\(f^{'}(x)=2x-1\)
\(f^{'}(0)=-1\)
then the tangent slope is\(\boxed{ -1}\)

since the tangent is parallel to the x-axis
then his Slope =0
assume that the tangent point  equal \((a,f(a)) \)
since the slope =0
then \[f^{'}(a)=0\]
\[f^{'}(x)=3x^{2}-6x\]
but \[f^{'}(a)=0\]
then \[3a^{2}-6a=0\]
\[3a(a-2)=0\]
\[a=0 \ or \ a=2\]
the points are \((0,f(0)),(2,f(2))\)

from vertical line equation
the vertical line slope=\(\frac{1}{5}\)
then the tangent slope =\(-5\)
* to find the tangent point
\(f^{'}(x)=-5\)
\(f^{'}(x)=2x+1=-5\)
\(\boxed{x=-3}----- the \ X-coordinate\)
Now substite \(\boxed{x=-3}\) in \(f(x)\) to get the Y-coordinate
the the tangent point is \((-3,1)\)
the tangent equation is \(y-1=-5(x-(-3))\)
\(\boxed{y=-5x-14} ---> \ tangent \ equation\)