# Revision history [back]

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

var('x')

y = x^4-3x+8

solve(diff(y,x), x)

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

var('x')

y = x^4-3x+8x^4-3*x+8

solve(diff(y,x), x)

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

var('x')

y f = x^4-3*x+81/4log(2.02000000000000x + 1) + 1/2log(0.00999999999999979x + 1) + 1/4log(-2x + 1)

solve(diff(y,x), x)find_local_maximum(diff(f,x), 0, 1)

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

var('x')

f = 1/4log(2.02log(2.02000000000000x + 1) / 2 + log(-2x + 1) + 1/2log(0.00999999999999979x + 1) + 1/4log(-2x + 1)/ 2

find_local_maximum(diff(f,x), 0, 1)

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

f = log(2.02log(2.02 * x + 1) / 2 + log(-2log( -2 * x + 1) / 2

find_local_maximum(diff(f,x), 0, 1)

What is mysterious for me, is that the above works if instead I have

f = log( 1.02 * x + 1) / 2 + log( -x + 1) / 2

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

f = log(2.02 * x + 1) / 2 + log( -2 * x + 1) / 2

find_local_maximum(diff(f,x), 0, 1)

What is mysterious for me, is that the above works if instead I have

f = log( 1.02 log(1.02 * x + 1) / 2 + log( -x + 1) / 2

### Numerical real solution of derivative

I would like to know where a function attains its maximum, so I'm trying to solve some

diff(y,x),x

where y depends on y. I have difficulties with Sage returning equations, complex roots, converting equations, find_maximum_on_interval command etc, and instead of spending another hour trying to figure it out myself, I thought I would try asking here...

Here is a more concrete example:

f = log(2.02 * x + 1) / 2 + log( -2 * x + 1) / 2

find_local_maximum(diff(f,x), 0, 1)

What is mysterious for me, is that the above works if instead I have

f = log(1.02 log(1.01 * x + 1) / 2 + log( -x + 1) / 2