Suppose that f(0) = -3 and f'(x) le 5 for all | vedclass

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(A) According to the Lagrange's Mean Value Theorem $(LMVT)$,for a function $f$ continuous on $[0, 2]$ and differentiable on $(0, 2)$,there exists at l

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$7$

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(A) According to the Lagrange's Mean Value Theorem $(LMVT)$,for a function $f$ continuous on $[0, 2]$ and differentiable on $(0, 2)$,there exists at least one $c \in (0, 2)$ such that:$\frac{f(2) - f(0)}{2 - 0} = f'(c)$Given that $f(0) = -3$ and $f'(x) \le 5$ for all $x$,we substitute these into the equation:$\frac{f(2) - (-3)}{2} = f'(c)$$\frac{f(2) + 3}{2} = f'(c)$Since $f'(c) \le 5$,we have:$\frac{f(2) + 3}{2} \le 5$$f(2) + 3 \le 10$$f(2) \le 7$Therefore,the largest value $f(2)$ can attain is $7$.

Suppose that $f(0) = -3$ and $f'(x) \le 5$ for all values of $x$. Then the largest value which $f(2)$ can attain is

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