If f is twice differentiable such that f''(x) | vedclass

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(C) Given $h'(x) = [f(x)]^2 + [g(x)]^2$. Taking the derivative with respect to $x$: $h''(x) = 2f(x)f'(x) + 2g(x)g'(x)$. Since $f'(x) = g(x)$,we have $

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a straight line with slope $2$

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(C) Given $h'(x) = [f(x)]^2 + [g(x)]^2$. Taking the derivative with respect to $x$: $h''(x) = 2f(x)f'(x) + 2g(x)g'(x)$. Since $f'(x) = g(x)$,we have $g'(x) = f''(x) = -f(x)$. Substituting these into the expression for $h''(x)$: $h''(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$. Since $h''(x) = 0$,$h'(x)$ must be a constant,say $C$. Thus,$h(x) = Cx + D$. Given $h(0) = 2$,we get $D = 2$. Given $h(1) = 4$,we get $C(1) + 2 = 4$,which implies $C = 2$. Therefore,$h(x) = 2x + 2$. This is a straight line with slope $2$ and $y$-intercept $2$.

If $f$ is twice differentiable such that $f''(x) = -f(x)$,$f'(x) = g(x)$,$h'(x) = [f(x)]^2 + [g(x)]^2$,and $h(0) = 2$,$h(1) = 4$,then the equation $y = h(x)$ represents:

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