Difference between revisions of "009A Sample Midterm 3, Problem 3"

Latest revision as of 13:22, 11 November 2017

Let $y=3{\sqrt {2x+5}},x\geq 0.$

(a) Use the definition of the derivative to compute ${\frac {dy}{dx}}.$

(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=3\sqrt{2x+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 (2,9).}

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam"> Use the definition of the derivative to compute &nbsp; <math>\frac{dy}{dx}</math> &nbsp; for &nbsp;<math style="vertical-align: -4px">y=3\sqrt{-2x+5}.</math>
+<span class="exam"> Let &nbsp;<math style="vertical-align: -3px">y=3\sqrt{2x+5},x\ge 0.</math>
+<span class="exam">(a) Use the definition of the derivative to compute &nbsp; <math style="vertical-align: -13px">\frac{dy}{dx}.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<span class="exam">(b) Find the equation of the tangent line to &nbsp;<math style="vertical-align: -3px">y=3\sqrt{2x+5}</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">(2,9).</math>
-!Foundations: &nbsp;
+<hr>
-|-
-|Recall
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
-|}
+[[009A Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Step 1: &nbsp;
-|-
-|Let &nbsp;<math style="vertical-align: -5px">f(x)=3\sqrt{-2x+5}.</math>
-|-
-|Using the limit definition of the derivative, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
-&&\\
-& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2(x+h)+5}-3\sqrt{-2x+5}}{h}}\\
-&&\\
-& = & \displaystyle{\lim_{h\rightarrow 0} \frac{3\sqrt{-2x+-2h+5}-3\sqrt{-2x+5}}{h}}\\
-&&\\
-& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{\sqrt{-2x+-2h+5}-\sqrt{-2x+5}}{h}.}
-\end{array}</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we multiply the numerator and denominator by the conjugate of the numerator.
-|-
-|Hence, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(\sqrt{-2x+-2h+5}-\sqrt{-2x+5})}{h} \frac{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}{(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
-&&\\
-& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{(-2x+-2h+5)-(-2x+5)}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
-&&\\
-& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2h}{h(\sqrt{-2x+-2h+5}+\sqrt{-2x+5})}}\\
-&&\\
-& = & \displaystyle{3\lim_{h\rightarrow 0} \frac{-2}{\sqrt{-2x+-2h+5}+\sqrt{-2x+5}}}\\
-&&\\
-& = & \displaystyle{3\frac{-2}{\sqrt{-2x+5}+\sqrt{-2x+5}}}\\
-&&\\
-& = & \displaystyle{-\frac{3}{\sqrt{-2x+5}}.}
-\end{array}</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>-\frac{3}{\sqrt{-2x+5}}</math>
-|-
-|
-|}
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Midterm 3, Problem 3"

Latest revision as of 13:22, 11 November 2017

Navigation menu

Search