Equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.

Question

Let $f$ be differentiable function and $y=-2x+1$ be tangent line to graph of $y=f(x)$ at the point $x=2$.

How can I find the equation for tangent line to the graph $y=(f(x))^2$ at point $x=2$.

I created a function $f (x)$ myself, but it didn't work. Any ideas?

Answer 1

Find $f(2)$ and $f'(2)$ as follows $$\begin{align} y \quad &= \quad f(2) + f'(2)(x-2) \\&=\quad f'(2)x + f(2) - 2f'(2) \\&=\quad -2x + 1 \end{align}$$ Since the coefficients should be equal, we get $$\begin{cases} f'(2) = -2 \\ f(2) = -3 \end{cases}$$ Now, take $g = f^2$. To find the tangent equation, we will need the derivative of $g$ at $2$. $$g'(2) = [f(x)^2]'\bigg|_2 = 2f(2)f'(2) = 2(-3)(-2) = 12$$ Hence, the required tangent is $$\begin{align} y \quad &= \quad g(2) + g'(2)(x - 2) \\&=\quad (-3)^2 + 12(x-2) \tag{$g(2) = f(2)^2$} \\&=\quad 12x - 15 \end{align}$$

Answer 2

We know that the tangent line to a function $f$ at a point $(a, f(a))$ is given by the equation $$(y - f(a)) = f'(a) (x - a) \Leftrightarrow y = f'(a) x + (f(a) - f'(a) a).$$ Since $y = -2x + 1$ and $a = 2$, you find $f'(2) = -2$ and $f(2) = -3$. Then, you compute the tangent line to $g(x) = (f(x))^2$ at $a = 2$, $$(y - g(a)) = g'(a) (x - a).$$ The expression of $g'(2)$ is just an application of the chain rule.