Answer
$y=-x+2$
The graph of the function and the tangent line is shown below:
Work Step by Step
$y=2x-x^{3},$ at $(1,1)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=1$:
Find $f(a+h)$ by substituting $x$ by $1+h$ in $f(x)$ and simplifying:
$f(1+h)=2(1+h)-(1+h)^{3}=...$
$...=2+2h-[1+3(1)^{2}h+3(1)h^{2}+h^{3}]=...$
$...=2+2h-1-3h-3h^{2}-h^{3}=...$
$...=-h^{3}-3h^{2}-h+1$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=1$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{-h^{3}-3h^{2}-h+1-1}{h}=...$
$...=\lim_{h\to0}\dfrac{-h^{3}-3h^{2}-h}{h}=\lim_{h\to0}\dfrac{h(-h^{2}-3h-1)}{h}=...$
$...=\lim_{h\to0}(-h^{2}-3h-1)=-(0)^{2}-3(0)-1=-1$
The point-slope form of the equation of a line is $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are known. Substitute them into the formula to obtain the equation of the tangent line:
$y-1=-1(x-1)$
$y-1=-x+1$
$y=-x+1+1$
$y=-x+2$
