Question

Extending the ldeas Find the unique value of \(k\) that makes the function \(f(x)=\left\\{\begin{array}{ll}{x^{3},} & {x \leq 1} \\ {3 x+k,} & {x>1}\end{array}\right.\) differentiable at \(x=1 .\)

Short answer

For the given function, at \(x = 1\), the values of the derivatives from both the parts always remain equal, regardless of the value of 'k'. So, the function is differentiable at \(x = 1\) for any real number 'k'.

Step-by-step solution

Compute the derivative of the first segment

Since the first segment \(f(x) = x^3\) is a power function, its derivative with respect to 'x' can be found using the power rule for derivatives, which gives: \(f'(x) = 3x^2\).

Compute the derivative of the second segment

In the second segment \(f(x) = 3x + k\), the derivative of '3x' with respect to 'x' is '3', and since 'k' is a constant, its derivative is zero. Hence, \(f'(x) = 3\).

Set the derivatives equal at x = 1

Once the derivatives of both segments are obtained, the next step is to equate these derivatives at 'x = 1'. This gives \(3*(1)^2 = 3\). This equation simplifies to \(3 = 3\), confirming that the function is differentiable at x = 1, regardless of the value of 'k'.