The problem-solving process of freshman mathematics

Functions are continuous in the domain.

When 0

When x

When x->; 0-，f '(x)= 3x 2-2x = 0； When x->; 0+，f' (x) = 2x-3x 2 = 0。

So when x=0, the left and right derivatives are equal.

Similarly, when x= 1

x-& gt； 1-,f'(x)=2x-3x^2=- 1；

x-& gt； 1+，f' (x) = 3x 2-2x= 1。 At this time, the left and right derivatives of f(x) are not equal at x = 1 f( 1)=0

So the nondifferentiable point is (1, 0)

Let F(x)=f(x)-2, then F'(x)=f'(x). 0 F(x)/3x=5, we can get F(x)=0, that is, f(x)=2.

By the definition of derivative, f' (0) = f' (0) = limx->; 0 [F(0+x)-F(0)]/x

= limx-& gt； 0 F(x)/x= 15

So f'(0)= 15.

Let f(x)=y, obviously f (0) = 0;

And sin( 1/x) is bounded, | sin (1/x) | < = 1. So limx->; 0 xsin( 1/x)-& gt； 0

So the function is continuous.

And f' (x) = sin (1/x)-cos (1/x)/x, in x->; 0+，x-& gt； None of the zero derivatives exist.

So the function is continuous, but not differentiable at x=0.

y'= 1/(x+√( 1+x^2))*(x+√( 1+x^2))'

=[ 1+x/√( 1+x^2)]/(x+√( 1+x^2))

= 1/√( 1+x^2)

y''=[ 1/√( 1+x^2)]'

=[( 1+x^2)^(- 1/2)]'

=(- 1/2)*2x*( 1+x^2)^(-3/2)

=-x( 1+x^2)^(-3/2)