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 $f(x)=\frac{6x}{2+3x}$. 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.

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.

You can still ask an expert for help

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.

$m=\frac{\mathrm{\u25b3}y}{\mathrm{\u25b3}x}=\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$

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.

$m(x)=\frac{\frac{6}{5}-f(x)}{1-x}\text{}but\text{}since\text{}f(x)=\frac{6x}{2+3x}$ then the equation becomes.

$m(x)=\frac{\frac{6}{5}-\frac{6x}{2+3x}}{1-x}$ which can be but does not need tobe simplified to.

$m(x)=\frac{12}{15x+10}$ so just plug and chug now.

$m(1.1)\cong 0.45283$

$m(1.01)\cong 0.47714$

$m(1.001)\cong 0.47971$

$m(1.0001)\cong 0.47997$

$m(0.9)\cong 0.51064$

$m(0.99)\cong 0.48290$

$m(0.999)\cong 0.48029$

$m(0.9999)\cong 0.48003$

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 $\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$

$m=\frac{\mathrm{\u25b3}y}{\mathrm{\u25b3}x}=\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$

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.

$m(x)=\frac{\frac{6}{5}-f(x)}{1-x}\text{}but\text{}since\text{}f(x)=\frac{6x}{2+3x}$ then the equation becomes.

$m(x)=\frac{\frac{6}{5}-\frac{6x}{2+3x}}{1-x}$ which can be but does not need tobe simplified to.

$m(x)=\frac{12}{15x+10}$ so just plug and chug now.

$m(1.1)\cong 0.45283$

$m(1.01)\cong 0.47714$

$m(1.001)\cong 0.47971$

$m(1.0001)\cong 0.47997$

$m(0.9)\cong 0.51064$

$m(0.99)\cong 0.48290$

$m(0.999)\cong 0.48029$

$m(0.9999)\cong 0.48003$

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 $\frac{f({x}_{1})-f({x}_{2})}{{x}_{1}-{x}_{2}}$

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