Questions: For each part (f(x)) and (a) so that the expression is one of the limit definitions of (f^prime(a)). (a) Find the simplest (f(x)) and (a) so that: (f^prime(a)=lim x rightarrow 3 frac2^x-8x-3) (f(x)=) (function of (x)) (a=) (number) (b) Find the simplest (f(x)) and (a) so that: (f^prime(a)=lim h rightarrow 0 fraccos left(fracpi4+hright) sin left(fracpi4+hright)-frac12h) (f(x)=) (function of (x)) (a=) (number)

$For each part (f(x)) and (a) so that the expression is one of the limit definitions of (f^prime(a)). (a) Find the simplest (f(x)) and (a) so that: (f^prime(a)=lim x rightarrow 3 frac2^x-8x-3) (f(x)=) (function of (x)) (a=) (number) (b) Find the simplest (f(x)) and (a) so that: (f^prime(a)=lim h rightarrow 0 fraccos left(fracpi4+hright) sin left(fracpi4+hright)-frac12h) (f(x)=) (function of (x)) (a=) (number)$

Transcript text: For each part $f(x)$ and $a$ so that the expression is one of the limit definitions of $f^{\prime}(a)$. (a) Find the simplest $f(x)$ and $a$ so that: \[ \begin{array}{l} f^{\prime}(a)=\lim _{x \rightarrow 3} \frac{2^{x}-8}{x-3} \\ f(x)=\square \text { (function of } x \text { ) } \\ a=\square \text { (number) } \end{array} \] (b) Find the simplest $f(x)$ and $a$ so that: \[ \begin{array}{l} f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{4}+h\right) \sin \left(\frac{\pi}{4}+h\right)-\frac{1}{2}}{h} \\ f(x)=\square \text { (function of } x \text { ) } \\ a=\square \text { (number) } \end{array} \]

Solution

Solution Steps

Solution Approach

(a) The given limit expression is in the form of the definition of the derivative. We need to identify $ f(x) $ and $ a $ such that: \[ f^{\prime}(a)=\lim _{x \rightarrow 3} \frac{f(x)-f(3)}{x-3} \] By comparing, we can see that $ f(x) = 2^x $ and $ a = 3 $.

(b) Similarly, the given limit expression is in the form of the definition of the derivative. We need to identify $ f(x) $ and $ a $ such that: \[ f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \] By comparing, we can see that $ f(x) = \cos(x) \sin(x) $ and $ a = \frac{\pi}{4} $.

Step 1: Identify $ f(x) $ and $ a $ for Part (a)

Given the limit expression: \[ f^{\prime}(a)=\lim _{x \rightarrow 3} \frac{2^{x}-8}{x-3} \] We recognize this as the definition of the derivative: \[ f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \] By comparing, we identify: \[ f(x) = 2^x \] \[ a = 3 \]

Step 2: Compute $ f'(a) $ for Part (a)

To find $ f'(a) $, we differentiate $ f(x) $ and evaluate at $ x = 3 $: \[ f(x) = 2^x \] \[ f'(x) = 2^x \ln(2) \] \[ f'(3) = 2^3 \ln(2) = 8 \ln(2) \]

Step 3: Identify $ f(x) $ and $ a $ for Part (b)

Given the limit expression: \[ f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{4}+h\right) \sin \left(\frac{\pi}{4}+h\right)-\frac{1}{2}}{h} \] We recognize this as the definition of the derivative: \[ f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \] By comparing, we identify: \[ f(x) = \cos(x) \sin(x) \] \[ a = \frac{\pi}{4} \]

Step 4: Compute $ f'(a) $ for Part (b)

To find $ f'(a) $, we differentiate $ f(x) $ and evaluate at $ x = \frac{\pi}{4} $: \[ f(x) = \cos(x) \sin(x) \] Using the product rule: \[ f'(x) = \cos(x) \cos(x) - \sin(x) \sin(x) = \cos^2(x) - \sin^2(x) \] Evaluating at $ x = \frac{\pi}{4} $: \[ f'\left(\frac{\pi}{4}\right) = \cos^2\left(\frac{\pi}{4}\right) - \sin^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 - \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2} - \frac{1}{2} = 0 \]

Final Answer

For Part (a): \[ f(x) = 2^x \] \[ a = 3 \] \[ f'(3) = 8 \ln(2) \] \[ \boxed{f(x) = 2^x, \, a = 3, \, f'(3) = 8 \ln(2)} \]

For Part (b): \[ f(x) = \cos(x) \sin(x) \] \[ a = \frac{\pi}{4} \] \[ f'\left(\frac{\pi}{4}\right) = 0 \] \[ \boxed{f(x) = \cos(x) \sin(x), \, a = \frac{\pi}{4}, \, f'\left(\frac{\pi}{4}\right) = 0} \]

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