The graph of the function y=x^3 + 6x^2 + 7x - 2cosx changes concavity at x=....?
We're supposed to use a calculator and the answer is -1.89 but how do you find that?
$\displaystyle y=x^3+6x^2+7x-2\cos x$
$\displaystyle y'=3x^2+12x+7+2\sin x$
$\displaystyle y''=6x+12+2\cos x$
Solve,
$\displaystyle 2\cos x+6x+12=0$
$\displaystyle \cos x+3x+4=0$
You need to use Newton's method.
Graph the function and see approximate value. Shown Below.
We need to find zero for the function,
$\displaystyle f(x)=\cos x+3x+4$
$\displaystyle f'(x)=-\sin x+3$
We note that,
$\displaystyle x\approx -1$
Define the iterative sequence,
$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
$\displaystyle x_{n+1}=x_n-\frac{\cos x_n+3x_n+4}{3-\sin x_n}$
$\displaystyle x_{n+1}=\frac{3x_n-x_n\sin x_n-\cos x_n-3x_n-4}{3-\sin x_n}$
$\displaystyle x_{n+1}=\frac{x_n\sin x_n+\cos x_n+4}{\sin x_n-3}$
The the iterative sequnce is,
$\displaystyle -1,-1.401,-1.3925,-1.3925,-1.3925,...$
Thus, the solution is approximately,
$\displaystyle x\approx -1.3925$
PH, I think you have a small typo in your derivative. Should be
$\displaystyle cos(x)+3x+6$. You have $\displaystyle cos(x)+3x+4$
The solution is, indeed, -1.89. I ran it through the ol' TI-92 to make doubly sure.
You may use the Intermediate Value Theorem. You can try x=-1.88 and get 0.055699
x=-1.9 and get -0.0232895.
Note the sign change. Therefore, you know it's between -1.88 and -1.9. Whittle away.