Let
f ( x ) = 1 x − 2 \displaystyle {f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}-{2}}}} f ( x ) = x − 2 1
Use the limit definition of the derivative to find
(i)
f ′ ( − 1 ) = \displaystyle {f}'{\left(-{1}\right)}= f ′ ( − 1 ) = Preview Question 6 Part 1 of 4 (ii)
f ′ ( 1 ) = \displaystyle {f}'{\left({1}\right)}= f ′ ( 1 ) = Preview Question 6 Part 2 of 4 (iii)
f ′ ( 3 ) = \displaystyle {f}'{\left({3}\right)}= f ′ ( 3 ) = Preview Question 6 Part 3 of 4 (iv)
f ′ ( 5 ) = \displaystyle {f}'{\left({5}\right)}= f ′ ( 5 ) = Preview Question 6 Part 4 of 4
To avoid calculating four separate limits, I suggest that you evaluate the derivative at the point when
x = a \displaystyle {x}={a} x = a . Once you have the derivative, you can just plug in those four values for "a" to get the answers.
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Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)