1. ## concavity

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?

2. Find the 2nd derivative, set to 0 and solve for x.

$\frac{d^{2}}{dx^{2}}[x^{3}+6x^{2}+7x-2cos(x)]=2cos(x)+6x+12$

Solve $2cos(x)+6x+12=0$ for x.

3. Originally Posted by abcocoa
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?
$y=x^3+6x^2+7x-2\cos x$
$y'=3x^2+12x+7+2\sin x$
$y''=6x+12+2\cos x$
Solve,
$2\cos x+6x+12=0$
$\cos x+3x+4=0$

You need to use Newton's method.
Graph the function and see approximate value. Shown Below.
We need to find zero for the function,
$f(x)=\cos x+3x+4$
$f'(x)=-\sin x+3$
We note that,
$x\approx -1$
Define the iterative sequence,
$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$
$x_{n+1}=x_n-\frac{\cos x_n+3x_n+4}{3-\sin x_n}$
$x_{n+1}=\frac{3x_n-x_n\sin x_n-\cos x_n-3x_n-4}{3-\sin x_n}$
$x_{n+1}=\frac{x_n\sin x_n+\cos x_n+4}{\sin x_n-3}$
The the iterative sequnce is,
$-1,-1.401,-1.3925,-1.3925,-1.3925,...$
Thus, the solution is approximately,
$x\approx -1.3925$

4. ## but....

the answer is -1.89 not -1.39...

Also my problem is solving 6x+12+2cosx=0. How do i solve this?

5. Originally Posted by abcocoa
the answer is -1.89 not -1.39...
The answer you gave it wrong, just substituet it into the equation and see that it is not zero.
Also my problem is solving 6x+12+2cosx=0. How do i solve this?
I posted the solution above.

6. PH, I think you have a small typo in your derivative. Should be

$cos(x)+3x+6$. You have $cos(x)+3x+4$

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.