Mathematics - Calculus
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I really need help on these 4 problems and if you could please post your steps on how to do them, I'd greatly appreciate it.
#1 Find an equation for the line tangent to the curve at the point defined by the given value of t.
Also, find the value of d2y/dx2 at this point x=6t, y = √t, t = 1/25
The equation for the line tangent to the curve at t=1/25 is?
d2y/dx2| t= 1/25 = ?
#2. Find the function's absolute maximum and minimum values on the given interval.
F(x) = x 5/3, -8<x<1
The absolute maximum of the function f(x) = x5/3 on the interval -8<x<1 has a value of ?
The absolute minimum of the function f(x) = x5/3 on the interval -8<x<1 has a value of?
#3. Find the absolute maximum and minimum values of the function on the given interval. F(x) = 1/3 x+5, -1 < x < 3
The absolute maximum of the function f(x) = 1/3 x+5 on the interval -1<x<3 has a value of?
The absolute minimum of the function (x)=1/3x+5 on the interval -1<x<3 has a value of ?
#4 Find the value or values of c that satisfy the equation f(b)-f(a)/b-a = f(c) in the conclusion of the Mean Value Theorem for the following function and interval.
F(x) = 5x(squared) + 2x-3 [-1,0]
The values(s) of c that satisfy the equation f(b)-f(a)/b-a = f(c) is/are ?
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Find an equation for the line tangent to the curve at the point defined by the given value of t.
Also, find the value of d2y/dx2 at this point x=6t, y = √t, t = 1/25
The equation for the line tangent to the curve at t=1/25 is?
d2y/dx2| t= 1/25 = ?
(a) dx/dt = 6, dy/dt = 1/2√t
dy/dx = (dy/dt)/(dx/dt) = 1/(12√t) = 1/(12 √(1/25)) = 5/12
When t = 1/25, x = 6t = 6/25 and y = √t = √(1/25) = 1/5
The equation to the line ...
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