Identify the function’s extreme values in the given domain, and give the coordinates where they are located. Identify all local and all absolute maxima and minima. All work steps on scrap, answers on test.
g(t)=t33-112t2+28t,0=t<>
local minimum:
absolute minimum:
local maximum:
absolute maximim:
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The Correct Answer and Explanation is :
To solve this problem, we need to identify the local and absolute extreme values of the function ( g(t) = t^3 – 112t^2 + 28t ) within the given domain ( 0 \leq t \leq ) (it’s likely that the upper bound for ( t ) is missing).
Step 1: Find the first derivative
We begin by finding the first derivative of the function to locate the critical points, which occur where the first derivative equals zero or is undefined.
[
g'(t) = \frac{d}{dt}(t^3 – 112t^2 + 28t) = 3t^2 – 224t + 28
]
Step 2: Find the critical points
Set the first derivative equal to zero to find the critical points:
[
3t^2 – 224t + 28 = 0
]
We solve this quadratic equation using the quadratic formula:
[
t = \frac{-(-224) \pm \sqrt{(-224)^2 – 4(3)(28)}}{2(3)}
]
[
t = \frac{224 \pm \sqrt{50176 – 336}}{6}
]
[
t = \frac{224 \pm \sqrt{49840}}{6}
]
[
t = \frac{224 \pm 223.18}{6}
]
This gives us two critical points:
[
t_1 = \frac{224 + 223.18}{6} \approx 74.53 \quad \text{and} \quad t_2 = \frac{224 – 223.18}{6} \approx 0.14
]
Step 3: Determine the second derivative
Now, find the second derivative to determine whether the critical points are maxima or minima.
[
g”(t) = \frac{d}{dt}(3t^2 – 224t + 28) = 6t – 224
]
Step 4: Test the critical points
To classify the critical points, substitute ( t_1 ) and ( t_2 ) into the second derivative:
- For ( t_1 \approx 74.53 ):
[
g”(74.53) = 6(74.53) – 224 \approx 447.18 – 224 = 223.18 \quad (\text{positive, local minimum})
] - For ( t_2 \approx 0.14 ):
[
g”(0.14) = 6(0.14) – 224 \approx 0.84 – 224 = -223.16 \quad (\text{negative, local maximum})
]
Step 5: Evaluate function at critical points and endpoints
Next, evaluate the function at the critical points and the endpoints ( t = 0 ) and ( t = 74.53 ) (if this is the domain boundary):
- ( g(0) = 0^3 – 112(0^2) + 28(0) = 0 )
- ( g(74.53) \approx (74.53)^3 – 112(74.53)^2 + 28(74.53) )
[
g(74.53) \approx 411926.55 – 612497.97 + 2087.84 \approx -199483.58
] - ( g(0.14) \approx (0.14)^3 – 112(0.14)^2 + 28(0.14) )
[
g(0.14) \approx 0.002744 – 2.2208 + 3.92 \approx 1.701
]
Step 6: Conclusion
- The local maximum occurs at ( t \approx 0.14 ), with a value of ( g(0.14) \approx 1.701 ).
- The local minimum occurs at ( t \approx 74.53 ), with a value of ( g(74.53) \approx -199483.58 ).
- The absolute maximum is at ( t = 0 ), with a value of ( g(0) = 0 ).
- The absolute minimum is at ( t \approx 74.53 ), with a value of ( g(74.53) \approx -199483.58 ).