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| − | <span class="exam">Let <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math> | + | <span class="exam">Consider the following function <math style="vertical-align: -5px"> f:</math> |
| | + | ::<math>f(x) = \left\{ |
| | + | \begin{array}{lr} |
| | + | x^2 & \text{if }x < 1\\ |
| | + | \sqrt{x} & \text{if }x \geq 1 |
| | + | \end{array} |
| | + | \right. |
| | + | </math> |
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| − | <span class="exam">(a) Use the definition of the derivative to compute <math>\frac{dy}{dx}</math> for <math style="vertical-align: -5px">y=\sqrt{3x-5}.</math> | + | <span class="exam">(a) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math> |
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| − | <span class="exam">(b) Find the equation of the tangent line to <math style="vertical-align: -5px">y=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math> | + | <span class="exam">(b) Find <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math> |
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| | + | <span class="exam">(c) Find <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(d) Is <math style="vertical-align: -5px">f</math> continuous at <math style="vertical-align: -1px">x=1?</math> Briefly explain. |
| − | !Foundations:
| + | <hr> |
| − | |-
| + | [[009A Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']] |
| − | |'''1.''' Recall
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| − | |-
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| − | | <math style="vertical-align: -13px">f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}</math>
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| − | |-
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| − | |'''2.''' The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
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| − | |-
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| − | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math> | |
| − | |}
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| − | '''Solution:''' | + | [[009A Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |Let <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}.</math>
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| − | |-
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| − | |Using the limit definition of the derivative, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3(x+h)-5}-\sqrt{3x-5}}{h}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, we multiply the numerator and denominator by the conjugate of the numerator.
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| − | |-
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| − | |Hence, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(3x+3h-5)-(3x-5)}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3h}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{h\rightarrow 0} \frac{3}{\sqrt{3x+3h-5}+\sqrt{3x-5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{\sqrt{3x-5}+\sqrt{3x-5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{2\sqrt{3x-5}}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We start by finding the slope of the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1).</math>
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| − | |-
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| − | |Using the derivative calculated in part (a), the slope is
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{m} & = & \displaystyle{f'(2)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{2\sqrt{3(2)-5}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{2}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Now, the tangent line to <math style="vertical-align: -5px">f(x)=\sqrt{3x-5}</math> at <math style="vertical-align: -5px">(2,1)</math>
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| − | |-
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| − | |has slope <math style="vertical-align: -13px">m=\frac{3}{2}</math> and passes through the point <math style="vertical-align: -5px">(2,1).</math>
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| − | |-
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| − | |Hence, the equation of this line is
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| − | |-
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| − | | <math>y=\frac{3}{2}(x-2)+1.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>\frac{dy}{dx}=\frac{3}{2\sqrt{3x-5}}</math>
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| − | |-
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| − | | '''(b)''' <math>y=\frac{3}{2}(x-2)+1</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
