approximate each real zero

Question

Given f(x)=6x^4-7x^3-23x^2+14x+3, approximate each real zero as a decimal to the nearest tenth.

Answer 1

Another solution is to use Newton's Method.  If you know calculus then you can use this method to find the roots without a calculator.

 

X_n+1=X_n-f(X_n)/f'(X_n) where X₁ is a point that you choose that is close to where you think the root is.  It doesn't even have to be that close as long as the derivative isn't 0 and the point has a slope towards one of the roots.

f(X) is the original function and f'(X) is the derivative of the function.

 

After plugging in a point for X₁ it will give you X₂, etc. until you find each of the roots.  You will end up with the answers mentioned above, but this is how you solve for those without a calculator.

 

Answer 2

Dear Daniel,

 

Four real zeros were found:

 

x = -1.7

x = -0.17

x = 0.7

x = 2.3

 

I found these by graphing the equation and by using the TI-89 Titanium calculator.

 

I hope you are allowed to use calculators, Daniel.  Since the original "function" cannot be factored, I don't know how else one could arrive at an answer.

 

Answer 3

Simple investigation shows that f(-2)=35, f(-1)=-21, f(0)=3, f(1)=-7, f(2)=-21, f(3)=135. So you have roots between x=-2 and x=-1, x=-1 and x=0, x=0 and x=1, then fourth root is between x=2 and x=3. Use Newton's method for finding the roots.

 

x_n+1=x_n-f(x_n)/f'(x_n), where f'(x)=24x³-21x²-46x+14. x_n is the n^th approximation to the root. 

 

Example: Start with x=0. f(0)=3, f'(0)=14, therefore x₁=0-3/14=-3/14; x₂=(3/14)-f(3/14)/f'(3/14)≈=-0.17127; x₃=-0.17127-f(-0.17127)/f'(-0.17127)≈-0.16975. Since we need a root to the nearest tenth, x≈-0.2

 

 

If we repeat this starting from x=1, we will obtain x≈0.7

 

Notice, however, if you start with x=-1, you will get x=0.7 again!