If f(x) and g(x) are twice differentiable fun | vedclass

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(B) Given $f''(x) = g''(x)$, we have $f''(x) - g''(x) = 0$.Integrating both sides with respect to $x$, we get $f'(x) - g'(x) = c_1$.Given $f'(1) = 4$

Vedclass · Accepted Answer

$10$

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(B) Given $f''(x) = g''(x)$, we have $f''(x) - g''(x) = 0$.Integrating both sides with respect to $x$, we get $f'(x) - g'(x) = c_1$.Given $f'(1) = 4$ and $2g'(1) = 4 \implies g'(1) = 2$. Thus, $c_1 = 4 - 2 = 2$.So, $f'(x) - g'(x) = 2$.Integrating again, $f(x) - g(x) = 2x + c_2$.Given $f(2) = 9$ and $3g(2) = 9 \implies g(2) = 3$.Substituting $x = 2$ into the equation: $f(2) - g(2) = 2(2) + c_2 \implies 9 - 3 = 4 + c_2 \implies 6 = 4 + c_2 \implies c_2 = 2$.Thus, $f(x) - g(x) = 2x + 2$.At $x = 4$, $f(4) - g(4) = 2(4) + 2 = 8 + 2 = 10$.

If $f(x)$ and $g(x)$ are twice differentiable functions satisfying $f''(x) = g''(x)$, $f'(1) = 2g'(1) = 4$ and $f(2) = 3g(2) = 9$, then the value of $f(x) - g(x)$ at $x = 4$ is:

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