007A Sample Midterm 1 - Revision history

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Created page with "'''This is a sample, and is meant to represent the material usually covered in Math 7A for the midterm. An actual test may or may not be similar.''' '''Click on the''' '''<s..."

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'''This is a sample, and is meant to represent the material usually covered in Math 7A for the midterm. An actual test may or may not be similar.'''

'''Click on the''' '''<span class="biglink" style="color:darkblue;"> boxed problem numbers </span> to go to a solution.'''
<div class="noautonum">TOC</div>

== [[007A_Sample Midterm 1,_Problem_1|<span class="biglink"><span style="font-size:80%"> Problem 1 </span></span>]] ==
<span class="exam"> Find the following limits:

<span class="exam">(a) Find  <math style="vertical-align: -13px">\lim _{x\rightarrow 2} g(x),</math>  provided that  <math style="vertical-align: -15px">\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]=5.</math>

<span class="exam">(b) Find  <math style="vertical-align: -14px">\lim _{x\rightarrow 0} \frac{\sin(4x)}{5x} </math>

<span class="exam">(c) Evaluate  <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math>

== [[007A_Sample Midterm 1,_Problem_2|<span class="biglink"><span style="font-size:80%"> Problem 2 </span>]] ==
<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>

<span class="exam">(a) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^-} f(x).</math>

<span class="exam">(b) Find  <math style="vertical-align: -15px"> \lim_{x\rightarrow 1^+} f(x).</math>

<span class="exam">(c) Find  <math style="vertical-align: -13px"> \lim_{x\rightarrow 1} f(x).</math>

<span class="exam">(d) Is  <math style="vertical-align: -5px">f</math>  continuous at  <math style="vertical-align: -1px">x=1?</math>  Briefly explain.

== [[007A_Sample Midterm 1,_Problem_3|<span class="biglink"><span style="font-size:80%"> Problem 3 </span>]] ==
<span class="exam"> Let  <math style="vertical-align: -5px">y=2x^2-3x+1.</math>

<span class="exam">(a) Use the '''definition of the derivative''' to compute   <math style="vertical-align: -13px">\frac{dy}{dx}.</math>

<span class="exam">(b) Find the equation of the tangent line to  <math style="vertical-align: -4px">y=2x^2-3x+1</math>  at  <math style="vertical-align: -4px">(2,3).</math>

== [[007A_Sample Midterm 1,_Problem_4|<span class="biglink"><span style="font-size:80%"> Problem 4 </span>]] ==
<span class="exam"> Find the derivatives of the following functions. '''Do not simplify.'''

<span class="exam">(a)   <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>

<span class="exam">(b)   <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>

<span class="exam">(c)   <math style="vertical-align: -8px">h(x)=\sqrt{x+\sqrt{x}}</math>

== [[007A_Sample Midterm 1,_Problem_5|<span class="biglink"><span style="font-size:80%"> Problem 5 </span>]] ==
<span class="exam"> To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula:

::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>

<span class="exam">where  <math style="vertical-align: 0px">h</math>  is the height in centimeters and  <math style="vertical-align: 0px">m</math>  is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of  <math style="vertical-align: 0px">m=85.</math>  What is the rate at  <math style="vertical-align: -1px">h=170</math>  and  <math style="vertical-align: -1px">h=190?</math>  Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?

'''Contributions to this page were made by [[Contributors|Kayla Murray]]'''