1. find the limit and show work lim as x approaches 5 (abs value of (x-5))/(x-5)
ID: 3343745 • Letter: 1
Question
1. find the limit and show work
lim as x approaches 5 (abs value of (x-5))/(x-5)
2. s(t)= 96-32t^2 is the function which descrives how far an object falls in t sec from a height of 96 ft. Using limits, determine the object's velocity (instantaneous) at 1 sec.
3. find the derivative of function and simplify
f(x) - ((x^2)/3) + (4/(x^5))
4. Determine f '' (x) for f(x) 4x^3e^5
please show work for both derivatives
5. write an equation of a tangent line to f(x) = x/(x-7) at x=5. leave answer in slope intercept form.
6. On planet x, a ball dropped from a height of 50 m falls according to the equation s(t)=50-2.5t^2 . What is the acceleration due to gravity on this planet?
7. Differentiate (find dx/dy) don't simplify
y=(4x+3)^5 (x+1)^-2
Explanation / Answer
1.
lim x-->5 (|x-5|)/(x-5)
for x >5, |x-5| = x-5
so, lim x-->5++ [(x-5)/(x-5)] = 1
for x <5, |x-5| = -1
so, lim x-->5- [ -(x-5)/(x-5)] = -1
here lim x--> 5- doe not equal to lim x-->5+
so, the limit does not exist.
2.
v(t) = lim h-->0 [ 96-32(t+h)^2 - 96+32t^2]/h
= lim h-->0 [-32t^2-32h^2 - 64th +32t^2]/h
= lim h-->0 [-32h^2 - 64th ]/h
= lim h-->0 -32h - 64t
= -64t
at t = 1 sec
v(t) = -64*1 = -64 m/s
3.
f(x) = ((x^2)/3) + (4/(x^5))
differentiate
f'(x) = (2/3)*x - 20/x^6
= [2x^7 - 60]/[3x^6]
4. f(x) = 4x^3e^5x
differentiate
f'(x) = 4 * [x^3]' *e^5x + 4*x^3 * [e^5x]'
= 4 * 3x^2*e^5x + 4*x^3 * 5*e^(5x)
= 12*x^2 *e^5x + 20*x^3*e^(5x)
again differentiate
f''(x) = 12*[x^2]'*e^5x + 12*x^2*[e^(5x)]' + 20*[x^3]' *e^(5x) + 20*x^3*[e^(5x)]'
= 12*2*x *e^5x + 12*x^2*5*e^(5x) + 20*3*x^2 *e^(5x) + 20*x^3*5*e^(5x)
= 24*x *e^5x + 60*x^2*e^(5x) + 60*x^2 *e^(5x) + 100*x^3*e^(5x)
= 100*x^3*e^(5x) + 120*x^2*e^(5x) + 24*x*e^(5x)
5. f(x) = x/(x-7) at x=5
differentiate
f'(x) = [ x-7 -x]/[x-7]^2 = [-7]/[x-7]^2
put x = 5
slope = [-7]/[5-7]^2 = -7/4
f(5) = 5/[5-7] = -5/2
equation of the tangent at (5, f(5)) = (5, -5/2) is
y + 5/2 = (-7/4) * (x -5)
==>
y = -7x/4 + 35/4 - 5/2
==>
y = -7x/4 + 25/4
it is in slope intercept form y = mx + c
here m = -7/4, c = 25/4
6.
acceleration = s''(t) = [0 - 5t]' = -5
7. y = (4x+3)^5 (x+1)^-2
differentiate it
dy/dx = [(4x+3)^5]' * (x+1)^-2 + (4x+3)^5 *[(x+1)^-2]'
= 20 *(4x+3)^4* (x+1)^-2 - 2*(4x+3)^5 *(x+1)^-3
so,
dx/dy = 1/dy/dx = 1/[20 *(4x+3)^4* (x+1)^-2 - 2*(4x+3)^5 *(x+1)^-3]
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