Question

find the values for c and d that make the function differentiable:f(x) = 5x^2 + c if x = or < than 3dx^2 - 3 if x > 3

Answer 1

9514 1404 393

Answer:

  c = -3; d = 5

Step-by-step explanation:

The function is differentiable if it is continuous and the derivative is continuous.

This function will be continuous if the limit as x approaches 3 from either side is the same. From the left, the limit is ...

  5(3^2) +c = 45 +c

From the right, the limit is ...

  d(3^2) -3 = 9d -3

__

The derivative will be continuous if the limit as x approaches 3 from either side is the same. From the left, the limit is ...

  f'(x) = 10x

  = 10(3) = 30

From the right, the limit is ...

  f'(x) = 2dx

  = 2d(3) = 6d

__

This gives us a pair of simultaneous equations in 'c' and 'd':

  45 +c = 9d -3

  6d = 30

The latter tells us d = 5. Then the former tells us ...

  45 +c = 9(5) -3   ⇒   c = -3

The function is differentiable if c = -3 and d = 5.

_____

Additional comment

With these values, the function becomes ...

  $f(x)=\begin{cases}5x^2-3&\text{if }x\le 3\\5x^2-3&\text{if }x>3\end{cases}$