Question

Derivative help!
If f(x)=cos(pi/2 [x]-x^3),1 12 years ago

Answer 1

\[\Huge f(x)=\cos(\frac{\pi}{2}[x]-x^3),1<x<2\]

Answer 2

\[\LARGE and~[x]=GINT \le x ,then~f'(\sqrt[3]{\frac{\pi}{2}})=?\]

Answer 3

@oldrin.bataku @Amber101 @mathslover @terenzreignz @Rayvenn.14

Answer 4

@amistre64

Answer 5

Is [ ] referring to round, ceil, or floor function?

@DLS

Answer 6

floor

Answer 7

Notice that the floor function [x] for 1 < x < 2, will always be equal to 1 hence mimicking the horizontal line x = 1. So the derivative with the Chain Rule becomes:\[\bf f'(x)=3x^2\sin \left( \frac{\pi}{2}-x^3 \right)\]

Answer 8

Notice that the [x] function will always equal to 1 for all 1 < x < 2. Hence we can eliminate it from the function itself and differentiate normally with the Chain Rule.

Answer 9

yo!

Answer 10

Now that we know the derivative, evaluate the derivative at the given x-value:\[\bf f'\left( \sqrt[3]{\frac{\pi}{2}} \right)=3\left( \sqrt[3]{\frac{\pi}{2}} \right)^2\sin\left( \frac{\pi}{2}-\frac{\pi}{2} \right)=0\]

@DLS

Answer 11

@DLS Do you understand?

Answer 12

yes! thanks :D

Answer 13

$$f(x)=\cos\left(\frac\pi2\lfloor x\rfloor-x^3\right)$$Given \(1<x<2\), note that \(\lfloor x\rfloor=1\) and therefore tangent lines for the floor will be flat: