# Inc/dec/local min/max/concavity

• May 2nd 2010, 05:25 AM
SyNtHeSiS
Inc/dec/local min/max/concavity
11. f(x) = sinx + cosx, 0<= x <= 2pie

a) Find the intervals on which f is increasing or decreasing.
b) Find the local max and min values of f.
c) Find the intervals of concavity and the inflection points

I differentiated the equation to get f'(x) = cosx - sinx and then solved for the critical value which was pie / 4 but I dont know where do go from here.
• May 2nd 2010, 06:00 AM
nahduma
Quote:

Originally Posted by SyNtHeSiS
11. f(x) = sinx + cosx, 0<= x <= 2pie

a) Find the intervals on which f is increasing or decreasing.
b) Find the local max and min values of f.
c) Find the intervals of concavity and the inflection points

I differentiated the equation to get f'(x) = cosx - sinx and then solved for the critical value which was pie / 4 but I dont know where do go from here.

Actually, there are TWO critical values for $0\leq x\leq 2\pi :\frac{\pi}{4} and \frac{9\pi}{4}$
Now, the function is increasing if the slope is positive and vice versa.
For all $x \in [0,\frac{\pi}{4}] and x \in [\frac{9\pi}{4}, 2\pi] , f'(x) \geq 0. similarly, f'(x) \leq 0 \forall x \in [\frac{\pi}{4},\frac{9\pi}{4}]$.

I hope you can solve the rest of the problem from this:
the local minima and maxima are situated at the critical points.
if the second derivative is positive, the graph is convex. if it is negative, it is concave.
the point of inflection is where the second derivative changes signs (ie. from +ve to -ve or vice versa). therefore, it becomes zero.

EDIT:
Thanks to defunkt for pointing this out. 9pi/4 is obviously > 2pi. Read all occurrences as 5pi/4. (can't believe I was stupid enough to miss it) :(
• May 2nd 2010, 06:20 AM
Defunkt
Quote:

Originally Posted by nahduma
Actually, there are TWO critical values for $0\leq x\leq 2\pi :\frac{\pi}{4} and \frac{9\pi}{4}$
Now, the function is increasing if the slope is positive and vice versa.
For all $x \in [0,\frac{\pi}{4}] and x \in [\frac{9\pi}{4}, 2\pi] , f'(x) \geq 0. similarly, f'(x) \leq 0 \forall x \in [\frac{\pi}{4},\frac{9\pi}{4}]$.

I hope you can solve the rest of the problem from this:
the local minima and maxima are situated at the critical points.
if the second derivative is positive, the graph is convex. if it is negative, it is concave.
the point of inflection is where the second derivative changes signs (ie. from +ve to -ve or vice versa). therefore, it becomes zero.

Surely, you noted that $\frac{9\pi}{4} > \frac{8\pi}{4} = 2\pi$ ... (Wink)
• May 2nd 2010, 06:43 AM
nahduma
Actually, there are TWO critical values for $0\leq x\leq 2\pi :\frac{\pi}{4} and \frac{9\pi}{4}$