What is the answer for question 4. a) ?

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The given curve is `-(3x^3)/4+3x` .

(i) At `x=1` ,` f(x)=-(3*1^3)/4+3*1=-3/4+3=9/4=2.25`

At `x=2` , `f(x)=-(3*2^3)/4+3*2=-24/4+6=0`

**So, the slope of the secant between the points** (1,2.25) and (2,0) is

`(0-2.25)/(2-1)` =**-2.25.**

(ii) At `x=1` ,` f(x)=-(3*1^3)/4+3*1=-3/4+3=9/4=2.25`

At `x=1.5` , `f(x)=-(3*1.5^3)/4+3*1.5=-10.125/4+4.5=1.97`

**So, the slope of the secant between the points** (1,2.25) and (1.5,1.97) is

`(1.97-2.25)/(1.5-1)` =**-0.56**

(iii) At `x=1` ,` f(x)=-(3*1^3)/4+3*1=-3/4+3=9/4=2.25`

At` x=1.1` , `f(x)=-(3*1.1^3)/4+3*1.1=-3.993/4+3.3=2.302`

**So, the slope of the secant between the points** (1,2.25) and (1.1,2.302) is

`(2.302-2.25)/(1.1-1)` =**0.52**

(iv) At `x=1` ,` f(x)=-(3*1^3)/4+3*1=-3/4+3=9/4=2.25`

At `x=1.01` , `f(x)=-(3*1.01^3)/4+3*1.01=-3.09/4+3.03=2.2575`

**So, the slope of the secant between the points** (1,2.25) and (1.01,2.2575) is

`(2.2575-2.25)/(1.01-1)` =**0.75**

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