Question

Let f(x)={x^2-16x, x<9;12rootx, x> or equal to 9.Is f continuous at x=9? Determine whether f is differentiable at x=9.If so, find the value of the derivative there.

Answer 1

1.

$f(x) = \begin{cases} x^2 - 16x & \text{x<9} \\ 12 \sqrt{x} & \text{x $\ge 9$} \end{cases}$ . To determine whether f is continuous at x=9, we should check if: $\lim_{x \to 9-0} f(x) = \lim_{x \to 9+0} f(x) = f(9)$ , or in our terms: (since f differs when x<9 and x$\ge$9 ):

$\lim_{x \to 9-0} x^2 -16x = \lim_{x \to 9+0} 12 \sqrt{x} = f(9).$

Since $f(x) = 12 \sqrt{x} , x \ge 9,$ and $12 \sqrt{x}$ is continuous at every $x \in R$ , we obtain that $\lim_{x \to 9+0} 12 \sqrt{x} = f(9) = 36.$ So, we should check whether $\lim_{x \to 9-0} x^2 - 16x = 36$ . We will use the fact that $x^2 - 16x$ is continuous at every $x \in R$ , and thus at x=9: $\lim_{x->9-0}x^2 -16x = \lim_{x \to 9+0} x^2 -16x = 9^2 - 16 \cdot 9 = 81 - 144 = -63$ . So, we obtain that $\lim_{x \to 9-0} f(x) \neq \lim_{x \to 9+0} f(x)$. Hence, f is not continuous at x=9.

2.

Let's assume f is differentiable at x=9. We will use the fact that if f is differentiable at point $x_0$ then f is continuous at $x_0$ .

Hence, if f is differentiable at x=9, f should be contionuous at x=9, but we proved it is not. Thus, f can not be differentiable at x=9.