009A Sample Midterm 1, Problem 4 Detailed Solution
Let
(a) Use the definition of the derivative to compute for
(b) Find the equation of the tangent line to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=\sqrt{3x-5}} at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2,1).}
| Foundations: |
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| 1. Recall |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}} |
| 2. The equation of the tangent line to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} at the point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a,b)} is |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=m(x-a)+b} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=f'(a).} |
Solution:
(a)
| Step 1: |
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| Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sqrt{3x-5}.} |
| Using the limit definition of the derivative, we have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3(x+h)-5}-\sqrt{3x-5}}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}.} \end{array}} |
| Step 2: |
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| Now, we multiply the numerator and denominator by the conjugate of the numerator. |
| Hence, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\sqrt{3x+3h-5}-\sqrt{3x-5})}{h} \frac{(\sqrt{3x+3h-5}+\sqrt{3x-5})}{(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(3x+3h-5)-(3x-5)}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3}{\sqrt{3x+3h-5}+\sqrt{3x-5}}}\\ &&\\ & = & \displaystyle{\frac{3}{\sqrt{3x-5}+\sqrt{3x-5}}}\\ &&\\ & = & \displaystyle{\frac{3}{2\sqrt{3x-5}}.} \end{array}} |
(b)
| Step 1: |
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| We start by finding the slope of the tangent line to at |
| Using the derivative calculated in part (a), the slope is |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {\frac {3}{2{\sqrt {3(2)-5}}}}\\&&\\&=&\displaystyle {{\frac {3}{2}}.}\end{array}}} |
| Step 2: |
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| Now, the tangent line to at |
| has slope Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle m={\frac {3}{2}}} and passes through the point |
| Hence, the equation of this line is |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {3}{2}}(x-2)+1.} |
| Final Answer: |
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| (a) |
| (b) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {3}{2}}(x-2)+1} |