Let f(x)=x^3+6x^2+19x-1148.

Question

A solution to f(x)=0 can be found between x=8 and x=9.

The line tangent to the graph of y=f(x) at x=8 will have equation y=  

An approximate solution to f(x)=0 can be found at x1=   .

The line tangent to the graph of y=f(x) at x=x1 will have equation y= (Use the value of x1 you just found)

 

This suggests that an approximate solution to f(x)=0 can be found at x2=

Answer 1

Take f'(8) as 3(8)²+12(8)+19 or 307.  

Take f(8) as 8³+6(8)²+19(8)-1148 or -100.

Then construct -100 = 307x + b, where b is -100 - 307(8) or -2556.

The equation of the line tangent to the graph of 

y = x³+6x²+19x-1148 at (x,y) = (8,-100) is y = 307x - 2556.

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Take an approximate solution to f(x) = 0 at x₁ = 8.5; f(8.5) calculates to 61.125.

The line tangent to the graph of f(x) at (x,y) = (8.5,61.25) follows the form y = mx + b.

Place values and write 61.25 = [3(8.5)² + 12(8.5) + 19](8.5) + b where b amounts to -2809.625.

Then write y = 337.35x − 2809.625 as the tangent line to the graph of  y = x³+6x²+19x-1148 

at (x,y) = (8.5,61.25).

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For f(x) equal to x³+6x²+19x-1148 equated to 0, take f'(x) or dy/dx as

3x²+12x+19.

Now place "function over derivative" to obtain 

(x³+6x²+19x-1148)/(3x²+12x+19).

Next write x - (x³+6x²+19x-1148)/(3x²+12x+19)

and evaluate this expression for x = 8.5, exactly 

halfway between 8 & 9.

A programmable calculator gives an evaluation of 

8.319022946.

8.319022946 is then fed as "the new x" back into 

x - (x³+6x²+19x-1148)/(3x²+12x+19) to obtain 8.31588068.

Each new result of x - (x³+6x²+19x-1148)/(3x²+12x+19)

updates the value of x and [x - (x³+6x²+19x-1148)/(3x²+12x+19)]

is adjusted repeatedly for each change in x. 

The table below shows the progression of the value of x with its corresponding 

evaluation of [x - (x³+6x²+19x-1148)/(3x²+12x+19)].

x---------------------------[x - (x³+6x²+19x-1148)/(3x²+12x+19)]

8.5-------------------------------------------8.319022946

8.319022946------------------------------8.31588068

8.31588068--------------------------------8.315879743

8.315879743------------------------------8.315879743

In the last line of the table above, the evaluation of x - (x³+6x²+19x-1148)/(3x²+12x+19)

equals the last input value of x.

This signals an extremely accurate approximation of the root between 8 & 9 for x³+6x²+19x-1148 = 0.

As confirmation, x³+6x²+19x-1148 for x = 8.315879743 gives a result of -5.4E-8 (or -5.4 hundred-millionths) by programmable calculator, which might as well be treated as zero.



Answer 2

Graph used in the video:

desmos.com/calculator/j2dzq9sw1s

Newton's method:

desmos.com/calculator/0lxkq1v7c8