The point P(1, 6/5) lies on the graph of the function f defined by f(x)=(6x)/(2+3x). By calculating the slope of the secantline PQ for the successive points Q (x,f(x)) with x=1.1 x=1.01, x=1.001 x=1.0001 and x=0.9 x=0.99 x=0.999 x=0.9999. estimate (to 3 decimal places) the slope of the tangent lineto the graph of f at P.

Nash Frank 2022-07-31 Answered
The point P(1, 6/5) lies on the graph of the function f defined by f ( x ) = 6 x 2 + 3 x . By calculating the slope of the secantline PQ for the successive points Q (x,f(x)) with
x=1.1 x=1.01, x=1.001 x=1.0001
and
x=0.9 x=0.99 x=0.999 x=0.9999.
estimate (to 3 decimal places) the slope of the tangent lineto the graph of f at P.
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Answers (1)

i4epdp
Answered 2022-08-01 Author has 12 answers
The slope between the point P and any other point on the curve is found by.
m = y x = f ( x 1 ) f ( x 2 ) x 1 x 2
if we let x 1 = 1 (the x value of P) then f ( x 1 ) = 6 / 5 (the y value of P) and the slope ofany secant line from point P to any other point on the curve iscompletely defined by.
m ( x ) = 6 5 f ( x ) 1 x   b u t   s i n c e   f ( x ) = 6 x 2 + 3 x then the equation becomes.
m ( x ) = 6 5 6 x 2 + 3 x 1 x which can be but does not need tobe simplified to.
m ( x ) = 12 15 x + 10 so just plug and chug now.
m ( 1.1 ) 0.45283
m ( 1.01 ) 0.47714
m ( 1.001 ) 0.47971
m ( 1.0001 ) 0.47997
m ( 0.9 ) 0.51064
m ( 0.99 ) 0.48290
m ( 0.999 ) 0.48029
m ( 0.9999 ) 0.48003
Therefore an estimation of the slope of the tangent line of f atP to 3 decimal places is 0.480 sincethe limit of the slopes of the secant lines appears to beapproaching that number.
note that what I did above was just a little algebra to make lifeeasier, you could just as well plug into the general formula everytime if you like f ( x 1 ) f ( x 2 ) x 1 x 2
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