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On a test, I was marked wrong on the following 2 questions.
If you were the teacher, would you have marked me wrong?
1. State the definition of the 1st derivative.
My answer:
Function that gives the slope of the tangent line for any value of x that exists.
My score on this question: 0/2
The answer the teacher wanted:
[tex]
f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}
[/tex]
Obviously, although he asked for the definition, he wanted the mathematical definition, and gave no points to anyone who gave a word-based definition.
2. Suppose and object is thrown down from the top of a building at 100 ft/s. The building 1100 feet high. Its height s(t) after t seconds is given by s(t) = 1100-100t - 16t2.
a. Find the object's average velocity on the interval [1, 3]
What I did:
s'(t)=-100-32t
s'(1)=-100-32(1)=-132
s'(3)=-100-32(3)=-196
[tex]\frac{-132+(-136)}{2}=-164 ft/s[/tex]
My score 1.5/2
What he wanted:
[tex]
\frac{s(3)-s(1)}{3-1} =
\frac{1100-100(3)-16(3)^2-(1100-100(1)-16(1)^2)}{3-1}) = -164 ft/s
[/tex]
There were certain problems on this test where he told us that we may not use certain methods (i.e. no power rule allowed on certain problems), but he did not specify a method here.
If you were the teacher, would you have marked me wrong?
1. State the definition of the 1st derivative.
My answer:
Function that gives the slope of the tangent line for any value of x that exists.
My score on this question: 0/2
The answer the teacher wanted:
[tex]
f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}
[/tex]
Obviously, although he asked for the definition, he wanted the mathematical definition, and gave no points to anyone who gave a word-based definition.
2. Suppose and object is thrown down from the top of a building at 100 ft/s. The building 1100 feet high. Its height s(t) after t seconds is given by s(t) = 1100-100t - 16t2.
a. Find the object's average velocity on the interval [1, 3]
What I did:
s'(t)=-100-32t
s'(1)=-100-32(1)=-132
s'(3)=-100-32(3)=-196
[tex]\frac{-132+(-136)}{2}=-164 ft/s[/tex]
My score 1.5/2
What he wanted:
[tex]
\frac{s(3)-s(1)}{3-1} =
\frac{1100-100(3)-16(3)^2-(1100-100(1)-16(1)^2)}{3-1}) = -164 ft/s
[/tex]
There were certain problems on this test where he told us that we may not use certain methods (i.e. no power rule allowed on certain problems), but he did not specify a method here.
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