For each slope below,
enter the value correct to four decimal places.
Let
f ( x ) = 2 ln ( x ) \displaystyle {f{{\left({x}\right)}}}={2}{\ln{{\left({x}\right)}}} f ( x ) = 2 ln ( x ) . Using the formula,
m = f ( x 2 ) − f ( x 1 ) x 2 − x 1 \displaystyle {m}=\frac{{{f{{\left({x}_{{{2}}}\right)}}}-{f{{\left({x}_{{{1}}}\right)}}}}}{{{x}_{{{2}}}-{x}_{{{1}}}}} m = x 2 − x 1 f ( x 2 ) − f ( x 1 ) , find the slope of the secant line on the graph of
f ( x ) \displaystyle {f{{\left({x}\right)}}} f ( x ) that connects the points
x 1 = 9 \displaystyle {x}_{{{1}}}={9} x 1 = 9 and ...
...
x 2 = 14 \displaystyle {x}_{{{2}}}={14} x 2 = 14 :
m = \displaystyle {m}= m = Preview Question 6 Part 1 of 5
...
x 2 = 11 \displaystyle {x}_{{{2}}}={11} x 2 = 11 :
m = \displaystyle {m}= m = Preview Question 6 Part 2 of 5
...
x 2 = 10 \displaystyle {x}_{{{2}}}={10} x 2 = 10 :
m = \displaystyle {m}= m = Preview Question 6 Part 3 of 5
...
x 2 = 9.1 \displaystyle {x}_{{{2}}}={9.1} x 2 = 9.1 :
m = \displaystyle {m}= m = Preview Question 6 Part 4 of 5
...
x 2 = 9.01 \displaystyle {x}_{{{2}}}={9.01} x 2 = 9.01 :
m = \displaystyle {m}= m = Preview Question 6 Part 5 of 5
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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 DNE for Does Not Exist, oo for Infinity
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4) Enter DNE for Does Not Exist, oo for Infinity
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4) Enter DNE for Does Not Exist, oo for Infinity
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4) Enter DNE for Does Not Exist, oo for Infinity
Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4) Enter DNE for Does Not Exist, oo for Infinity