f(x)= ) 7x-8x^2&forxlt 0 7x^2+7x&forxgeqslant 0 According to the definition of the derivative, to compute f'(0) we need to compute the left-hand limit lim _(xarrow 0^-) which is square and the right-hand limit lim _(xarrow 0^+) which is square We conclude that f'(0) is square Note: If a limit or derivative is undefined, enter "undefined' as your answer.

To compute $f'(0)$, we need to compute the left-hand limit and the right-hand limit.<br /><br />The left-hand limit is the limit of the derivative of the function as $x$ approaches $0$ from the left side. In this case, the function is defined as $f(x) = 7x - 8x^2$ for $x < 0$. Taking the derivative of this function, we get $f'(x) = 7 - 16x$. Evaluating this at $x = 0$, we get $f'(0^-) = 7$.<br /><br />The right-hand limit is the limit of the derivative of the function as $x$ approaches $0$ from the right side. In this case, the function is defined as $f(x) = 7x^2 + 7x$ for $x \geq 0$. Taking the derivative of this function, we get $f'(x) = 14x + 7$. Evaluating this at $x = 0$, we get $f'(0^+) = 7$.<br /><br />Since the left-hand limit and the right-hand limit are both equal to $7$, we conclude that $f'(0) = 7$.