Suppose f(x) is differentiable at x = 1 and m | vedclass

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(A) Given that $f(x)$ is differentiable at $x = 1$,the derivative is defined as $f'(1) = \mathop {\lim }\limits_{h 	o 0} \frac{f(1 + h) - f(1)}{h}$.W

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$5$

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(A) Given that $f(x)$ is differentiable at $x = 1$,the derivative is defined as $f'(1) = \mathop {\lim }\limits_{h 	o 0} \frac{f(1 + h) - f(1)}{h}$.We are given $\mathop {\lim }\limits_{h 	o 0} \frac{f(1 + h)}{h} = 5$.For this limit to exist and be finite,the numerator $f(1 + h)$ must approach $0$ as $h 	o 0$,which implies $f(1) = 0$.Substituting $f(1) = 0$ into the definition of the derivative,we get:$f'(1) = \mathop {\lim }\limits_{h 	o 0} \frac{f(1 + h) - 0}{h} = \mathop {\lim }\limits_{h 	o 0} \frac{f(1 + h)}{h} = 5$.

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$	$II$. $f(x) = \frac{9}{28}x^{7/3} - 2$
$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$	$IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

Suppose $f(x)$ is differentiable at $x = 1$ and $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}f(1 + h) = 5$,then $f'(1)$ equals

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$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$ $II$. $f(x) = \frac{9}{28}x^{7/3} - 2$

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