Prove that f is continuous but not differentiable at the point (0, 0) .

Determine the value of $\int_{ - a}^a {f(x){\text{d}}x}$ where $a > 0$ .

we note that $f(0) = 0,{\text{ }}f(x) = 3x$ for $x > 0$ and $f(x) = x{\text{ for }}x < 0$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} x = 0$     M1A1

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 3x = 0$     A1

since $f(0) = 0$ , the function is continuous when x = 0     AG

$\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{h}{h} = 1$     M1A1

$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3h}}{h} = 3$     A1

these limits are unequal     R1

so f is not differentiable when x = 0     AG

[7 marks]

$\int_{ - a}^a {f(x){\text{d}}x = \int_{ - a}^0 {x{\text{d}}x + \int_0^a {3x{\text{d}}x} } }$     M1

$= \left[ {\frac{{{x^2}}}{2}} \right]_{ - a}^0 + \left[ {\frac{{3{x^2}}}{2}} \right]_0^a$     A1

$= {a^2}$     A1

[3 marks]