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Part 1 No Calculator
`lim_{x to 2} 3=`
`lim_{x to 5} x=`
`lim_{x to -2} (x^2+3x-4)=`
`lim_{x to 0+} 1/x=`
`lim_{x to 7} (x^2-49)/(x-7)=`
For what values of `x` is `f(x)={(3-(x-1)^2,,x < 1),(x+1,,x >= 1):}` continuous?
Use the limit definition of the derivative to find the slope of the tangent line to the curve `f(x)=x^2` at `x=-2`.
`f(-2+h)=`
`f(-2+h)-f(4)=`
`(f(-2+h)-f(4))/h=`
`lim_{h to 0}(f(-2+h)-f(-2))/h=`
So, `f'(-2)=`
Part 2 Calculator Allowed
Secants and Tangents
Estimate the slope of the tangent line (rate of change)
to `f(x)=1/x` at `x=2` by finding the slopes of the secant lines through the points:
`x=2.1` and `1.9`
`x=2.01` and `1.99`
Use the slopes of the secants to estimate the slope of the
tangent line accurate to 2 decimal places.
Sketch a graph of `f(x)` on the interval `[0,4]` along with
the tangent line when `x=2`.
Evaluate the following limits.
`lim_{x to 7} (x^2-49)/(x-7)=`
`lim_{x to -3} (2x^2+4x-5)=`
`lim_{x to 2^-} 1/(x-2)=`
`lim_{theta to 0} sin(4theta)/(4theta)=`
`lim_{x to 0} (1+2x)^(1/x)=`
`lim_{x to 0} (x-9)/(sqrt(x)-3)=`
`lim_{x to 0} (sqrt(3x+1)-1)/(2x)=`
`lim_{x to 0} (3/(x^2-x)+1/x)=`
If `f(x)={((x+3)^2-1,,x < -1),(2x-1,,x >= -1):}`
`lim_{x to -1^-}f(x)=`
`lim_{x to -1^+}f(x)=`
`lim_{x to -1}f(x)=`
Continuity
Determine whether `r(x)=(x^2-49)/(x-7)` is continuous at
`x=7`. If it is continuous, explain why. Otherwise, explain why
it is discontinuous.
Determine whether `p(x)=x^2-14x+49` is continuous at
`x=7`. If it is continuous, explain why. Otherwise, explain why
it is discontinuous.
Consider the function `f(x)=1/x` on the interval `-1<=x<=1`.
`f(-1)=-1` and `f(1)=1`, so, is there a value of `c` in `[-1,1]`
such that `f(c)=0`? Why or why not?
For what values of `x` is `g(x)=(x-5)/(x+2)` continuous?
The Precise Definition of a Limit
State the Precise Definition of a Limit.
If `f(x)=2x-1`, `a=-1`, and `epsilon=0.01`, then find an appropriate
value of `delta`.
If `f(x)=1/2 x+1` and `a=4`, then find `delta` in terms of `epsilon`.
The Derivative at a Point
Use the limit definition of the derivative to find the slope of the
tangent line to the curve `f(x)=5x^2` at `x=4`.
Evaluate each of the following and state your answers in simplest form:
`f(4+h)=`
`f(4+h)-f(4)=`
`(f(4+h)-f(4))/h=`
`lim_{h to 0}(f(4+h)-f(4))/h=`
So, `f'(4)=`
Use the limit definition of the derivative to find the slope of the
tangent line to the curve `f(x)=4x^2` at `x=1`.
`f(x)=sqrt(25-x)`. Use the limit definition of the derivative to compute `f'(9)`.
The Derivative Function
Use the limit definition of the derivative to find the
derivative function when `f(x)=5/x`.
Evaluate each of the following and state your answers in simplest form:
`f(x+h)=`
`f(x+h)-f(x)=`
`(f(x+h)-f(x))/h=`
`lim_{h to 0}(f(x+h)-f(x))/h=`
So, `f'(x)=`
Use the limit definition of the derivative to find the
derivative function when `f(x)=3x^2`.
`f(x)=sqrt(25-x)`. Use the limit definition of the derivative to
find the derivative function `f'(x)`.
`f(x)=x/(1-x^2)`. Use the limit definition of the derivative to
find the derivative function `f'(x)`.
`f(x)=1/x`. Use the limit definition of the derivative to
find the second derivative function `f''(x)`.