I'm tempted to "use calculus" to produce the Taylor Series for f(x) = sin(x) and then observe that:
\sin(x) < x - \frac{x^{3}}{6}
#3
Ackbach
Gold Member
MHB
4,148
93
I think I would approach it this way: minimize the function
$$f(x)=\sin(x)+2x-\frac{3x(x+1)}{\pi}$$
on the interval $[0,\pi/2]$ using the standard calculus techniques. There's one critical point near $0.88$, and then you have the endpoints. The left endpoint, I think, will end up being the smallest point on the graph. The second derivative might come in handy.
#4
awkward
364
0
Here is a slightly different approach.
Using f(x) as defined in post #3, show that \( f''(x) < 0 \) for all x, so f is concave. By computation, show that \( f(0) = 0 \) and \( f(\pi / 2) > 0 \). This is enough to conclude that \( f(x) > 0 \) on \( [0 , \pi/2] \).
Hi everybody
If we have not any answers for critical points after first partial derivatives equal to zero, how can we continue to find local MAX, local MIN and Saddle point?. For example: Suppose we have below equations for first partial derivatives:
∂ƒ/∂x = y + 5 , ∂ƒ/∂y = 2z , ∂ƒ/∂z = y
As you can see, for ∇ƒ= 0 , there are not any answers (undefined)