Calculus problems
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#1 Write an equation of the line tangent to the curve y=f(x) at the given point P on the curve. Express the answer in the form ax+by=c.
1)y=3x^2-4; P(1,-1)
2)y=2x-1/x; P(0.5,-1)
#2 Give the position function x=f(t) of a particle moving in a horizontal straight line. Find its location x when its velocity v is zero.
1)x=-16t^2=160t+25
#3 Give the height y(t) (in feet at time t seconds) of a ball thrown vertically upward. Find the maximum height that the ball attains.
1)y=-16t^2+128t+25
#4 Evaluate the Limits
1)lim as h goes to 0= 1/h(1/sqrt 9+h - 1/3)
2)lim x goes to 0= (sqrt 1+x - sqrt 1-x)/x
#5 Find a slope-predictor function for the given function f(x). Then write an equation for the line tangent to the curve y=f(x) at the point where x=2.
1)f(x)=x/x+1
2)f(x)=x^2+3/x
3)f(x)=x^2/x+1
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Solution Summary
Topics included in this solution set are: write an equation of a tangent line, give the position function of a particle, give the height of a ball thrown upwards, evaluate limits, and find a slope-predictor function.
Solution Preview
1.)In this case you need to calculate the derivative y'=f'(x) of each function. The derivative gives you the value of the slope of the line tangent to f(x) at point (x).
Once you have the derivative, you find the slope at the point you have been given. For instance in the first function you have to find f'(1) because 1 is the x coordinate of the point P you have been given. In the second case it would be f'(0.5).
Once you find the value of the slope you have all you need to calculate the equation of the line. You got a point P and a slope value.
The equation of a line is given by y=mx+b where m is the slope and x and y represent the coordinates of ...
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