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Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.001$, 0.01$, check this. I tested in all the other major CAS with the same both results ;)

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$)\right)=-\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.01$, check this. I tested in all the other major CAS with the same both results ;)

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=-\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.01$, check this. I tested in all the other major CAS with the same both results ;)

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=-\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.01$, check this. I tested in all the other major CAS with the same both results ;)

P.S.: the maximum in the interval $(0,1/2)$ is exactly $0.01$. It would be interesting to see how to get this answer directly. From mathematica I get something similar to sage, but maple and wolframalpha let me exactly the correct value. I will test geogebra now.

Ok, I see what is happening. The derivative of your function of the example is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=-\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.01$, check this. I tested in all the other major CAS with the same both results ;)

P.S.: the maximum in the interval $(0,1/2)$ is exactly $0.01$. It would be interesting to see how to get this answer directly. From mathematica I get something similar to sage, but maple and wolframalpha let me exactly the correct value. I will test geogebra now.

Ok, I see what is happening. The derivative of your function of the (the example that you put) is not bounded on the interval $(0,1)$, i.e. it is not defined in the points $-1/2$ and $1/2$ where it is going to infinity.

But sage works fine, it said that the maximum is in $1/2$ because it is an asymptote. In other words

$$\lim_{x\to\frac12^-}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=+\infty$$

and

$$\lim_{x\to\frac12^+}\frac{\mathrm d}{\mathrm dx}\left(\frac12(\log(2.02x + 1) + \log( -2x + 1) )\right)=-\infty$$

Check this:

http://www.wolframalpha.com/input/?i=limit+diff(log(2.02+*+x+%2B+1)+%2F+2+%2B+log(+-2+*+x+%2B+1)+%2F+2,x)+when+x-%3E1%2F2

So, Sage is not doing wrong... it is doing the "correct" answer. Well, the correct answer will be to say that there is no maximum in the interval, because the function is not bounded.

But if you change your interval by $(0,1/2)$ then you get the answer of $\approx 0.01$, check this. I tested in all the other major CAS with the same both results ;)

P.S.: the maximum in the interval $(0,1/2)$ is exactly $0.01$. It would be interesting to see how to get this answer directly. From mathematica I get something similar to sage, but maple and wolframalpha let me exactly the correct value. I will test geogebra now.

P.S2: geogebra found the exact value too.