9. $f(x)=\\begin{cases}cx + d&\\text{for }x\\leq2\\\\x^{2}-cx&\\text{for }x>2\\end{cases}$\nlet $f$ be the…

9. $f(x)=\\begin{cases}cx + d&\\text{for }x\\leq2\\\\x^{2}-cx&\\text{for }x>2\\end{cases}$\nlet $f$ be the function defined above, where $c$ and $d$ are constants. if $f$ is differentiable at $x = 2$, what\nis the value of $c + d$?

9. $f(x)=\\begin{cases}cx + d&\\text{for }x\\leq2\\\\x^{2}-cx&\\text{for }x>2\\end{cases}$\nlet $f$ be the function defined above, where $c$ and $d$ are constants. if $f$ is differentiable at $x = 2$, what\nis the value of $c + d$?

Answer

Explanation:

Step1: Continuity at (x = 2)

For the function (f(x)) to be continuous at (x = 2), (cx + d=x^{2}-cx) when (x = 2). Substitute (x = 2) into the equation: (2c + d=4 - 2c). Rearrange to get (d=4 - 4c).

Step2: Differentiability at (x = 2)

The derivative of (y=cx + d) is (y^\prime=c) (for (x\leq2)), and the derivative of (y=x^{2}-cx) is (y^\prime = 2x - c) (for (x>2)). For differentiability at (x = 2), (c=2(2)-c). Solve (c = 4 - c) for (c): Add (c) to both sides: (2c=4), so (c = 2).

Step3: Find (d)

Substitute (c = 2) into (d=4 - 4c). (d=4-4\times2=4 - 8=-4).

Step4: Calculate (c + d)

Substitute (c = 2) and (d=-4) into (c + d). (c + d=2+( - 4)=-2).

Answer:

B. -2