Question

\(f^{\prime}(x)= \begin{cases}-2 x, & x<0 \\ 2 x, & x \geq 0\end{cases}\) eqn of tangent $\rightarrow \quad y-y_{1}=2 x_{1}\left(x-x_{1}\right), x \geq 0$ \(y-y_{2}=-2 x_{2}\left(x-x_{2}\right) x<0\)

Short answer

Answer: For x < 0: $$ y - y_{2} = -2 x_{2}(x - x_{2}) $$ For x ≥ 0: $$ y - y_{1} = 2 x_{1}(x - x_{1}) $$

Step-by-step solution

Case 1: x < 0

In this case, the derivative function, f'(x), is given as -2x. Now we need to find the equation of the tangent line. To do this, consider an arbitrary point (x1, y1) with x1 < 0, since we are working with the case x < 0.

Equation of Tangent line for x < 0

Following the standard formula for the equation of a line, the equation of the tangent line when x < 0 will be: $$ y - y_{2} = -2 x_{2}(x - x_{2}), (x_{2} < 0) $$ This equation represents the tangent line for the given derivative function when x < 0.

Case 2: x ≥ 0

In this case, the derivative function, f'(x), is given as 2x. Now we need to find the equation of the tangent line. To do this, we consider an arbitrary point (x1, y1) with x1 ≥ 0.

Equation of Tangent line for x ≥ 0

Following the standard formula for the equation of a line, the equation of the tangent line when x ≥ 0 will be: $$ y - y_{1} = 2 x_{1}(x - x_{1}), (x_{1} \geq 0) $$ This equation represents the tangent line for the given derivative function when x ≥ 0. So, the equations of the tangent lines for the given piecewise derivative function are as follows: For x < 0: $$ y - y_{2} = -2 x_{2}(x - x_{2}) $$ For x ≥ 0: $$ y - y_{1} = 2 x_{1}(x - x_{1}) $$