# 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.

The point P(1, 6/5) lies on the graph of the function f defined by $f\left(x\right)=\frac{6x}{2+3x}$. 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.
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i4epdp
The slope between the point P and any other point on the curve is found by.
$m=\frac{\mathrm{△}y}{\mathrm{△}x}=\frac{f\left({x}_{1}\right)-f\left({x}_{2}\right)}{{x}_{1}-{x}_{2}}$
if we let ${x}_{1}=1$ (the x value of P) then $f\left({x}_{1}\right)=6/5$ (the y value of P) and the slope ofany secant line from point P to any other point on the curve iscompletely defined by.
then the equation becomes.
$m\left(x\right)=\frac{\frac{6}{5}-\frac{6x}{2+3x}}{1-x}$ which can be but does not need tobe simplified to.
$m\left(x\right)=\frac{12}{15x+10}$ so just plug and chug now.
$m\left(1.1\right)\cong 0.45283$
$m\left(1.01\right)\cong 0.47714$
$m\left(1.001\right)\cong 0.47971$
$m\left(1.0001\right)\cong 0.47997$
$m\left(0.9\right)\cong 0.51064$
$m\left(0.99\right)\cong 0.48290$
$m\left(0.999\right)\cong 0.48029$
$m\left(0.9999\right)\cong 0.48003$
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 $\frac{f\left({x}_{1}\right)-f\left({x}_{2}\right)}{{x}_{1}-{x}_{2}}$