I am stuck on a question on http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset3prb.pdf
3B-4
In the answer sheet, it says Both x2 and x3 are increasing functions on 0 ≤ x ≤ b.
I am wondering isn't it possible that b is smaller or equal to 0?

Question

I am stuck on a question on http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset3prb.pdf
3B-4
In the answer sheet, it says Both x2 and x3 are increasing functions on 0 ≤ x ≤ b.
I am wondering isn't it possible that b is smaller or equal to 0?

anonymous · Answer

Please help me, Thank you !

anonymous · Answer

The functions$$y=x^2$$and$$y=x^3$$are increasing within$$0 \le x \le +\infty$$,and decreasing within$$-\infty \le x \le \ 0$$.If any random point b is chosen on increasing part of the function, then function is defined (differentiated) within$$0 \le x \le b$$.If point$$x = b' = 0$$then you have$$x _{1}=x _{2}$$,and $$f(x _{1})=f(x _{2})$$,thus no the gradient, if b<0 then it is not defined on$$0 \le x \le b$$.