Calculus

#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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...a point in the line.
You already have a point P on the line, and you also have the value of the slope m, so all you have left is b, but you can calculate b by plugging the values you already have in your line equation. So in the first case where P=(x,y)=(1,-1) you get:
y=-1, m=f'(1), x=1. Plugging this in the equation you have:
-1 = f'(1)*1 + b
From this you can easily find b. Then the final equation for your line will be y=mx+b where m=f'(x) and b the value you just found.
2.)At this point you may already know that velocity is the derivative of position with respect to time. In this case you need to find the derivative of x with respect to t. ...