Let a function g : [ 0, 4 ] $\to$ R be defined as
$$g(x) = \left\{ {\matrix{
{\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr
{4 - x,} & {3 < x \le 4} \cr
} } \right.$$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
Answer (integer)
1
Solution
$f(x) = {x^3} - 6{x^2} + 9x - 3$<br><br>$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$<br><br>$f(1) = 1$, $f(3) = 3$<br><br>$$g(x) = \left[ {\matrix{
{f(9x)} & {0 \le x \le 1} \cr
0 & {1 \le x \le 3} \cr
{ - 1} & {3 < x \le 4} \cr
} } \right.$$<br><br>g(x) is continuous<br><br>$$g'(x) = \left[ {\matrix{
{3(x - 1)(x - 3)} & {0 \le x \le 1} \cr
0 & {1 \le x \le 3} \cr
{ - 1} & {3 < x \le 4} \cr
} } \right.$$<br><br>g(x) is non-differentiable at x = 3
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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