f(x)= ) -7x^2+5x&forxlt 0 6x^2-4&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 $\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}$ and the right-hand limit $\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}$.<br /><br />For the left-hand limit, we have $f(x) = -7x^2 + 5x$ for $x < 0$. Since $f(0)$ is undefined, we can't directly compute the limit. Therefore, the left-hand limit is undefined.<br /><br />For the right-hand limit, we have $f(x) = 6x^2 - 4$ for $x \geq 0$. Since $f(0)$ is defined and equal to $-4$, we can compute the limit as follows:<br /><br />$\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{6x^2 - 4 - (-4)}{x - 0} = \lim_{x \to 0^+} \frac{6x^2}{x} = \lim_{x \to 0^+} 6x = 0$<br /><br />Therefore, $f'(0)$ is 0.