Find any critical numbers of the function. (Enter your answers as a comma-separated list.)
g(t) = t sqrt6 − t , t < 13/3
I don't even know how to start out!
I will assume you meant
g(t) = t √(6-t) = t(6-t)^(1/2)
g'(t) = t (1/2)(6-t)^(-1/2) (-1) + (6-t)^(1/2) using the product rule
= 3)t-4)/(2√(6-t) )
=0 for any max/min values of the function
t/(2(6-t)^(1/2) = (6-t)^(1/2)
t = 2(6-t)
t = 12 - 2t
3t = 12
t = 4
so g(4) = 4√(6-4) = 4√2
to see if there are any points of inflection
differentiate 3(t-4)/(2√(6-t) ) , set that result equal to zero and solve for t
(hint: there are no points of intersection)
You should get a second derivative of 3(t-8)/(4(6-t)^(3/2 )
which when you set equal to zero will give an answer of
t = 8
BUT, t=8 is outside the domain of our function, so ... No Solution.
y-intercepts ? , let t =0
y = 0√6 = 0 , the origin (0,0)
any t-intercepts?
t(6-t)^(1/2) = 0
t = 0 or t = 6
look at the graph .....
http://www.wolframalpha.com/input/?i=t%286-t%29%5E%281%2F2%29
ignore the imaginary part of the graph (the red line)
the point (4,4√2) is a maximum point
To find the critical numbers of a function, you need to first find the derivative of the function and then set it equal to zero. In this case, let's find the derivative of the function g(t) = t√6 - t.
Step 1: Find the derivative of g(t) with respect to t.
To do this, you can apply the power rule and the constant rule. The power rule states that if f(x) = ax^n, then the derivative of f(x) with respect to x is f'(x) = nax^(n-1). The constant rule states that if f(x) = c, where c is a constant, then the derivative of f(x) with respect to x is f'(x) = 0.
Applying these rules, the derivative of g(t) is:
g'(t) = (1/2)√6*t^(1/2) - 1
Step 2: Set the derivative equal to zero and solve for t.
To find the critical numbers, we need to set g'(t) = 0 and solve for t.
(1/2)√6*t^(1/2) - 1 = 0
To isolate t, add 1 to both sides:
(1/2)√6*t^(1/2) = 1
Multiply both sides by 2/√6 to get rid of the fraction:
t^(1/2) = (2/√6)
Square both sides to eliminate the square root:
t = (2/√6)^2
Simplifying, we get:
t = (4/6)
t = 2/3
So, the critical number of the function g(t) = t√6 - t is t = 2/3.
Note: Since the given function is only defined for t < 13/3, it's important to check if the critical number t = 2/3 is within this range. In this case, 2/3 is indeed less than 13/3, so it is a valid critical number.