Mathematics Question Thread

[jamonwindeyer]

I'm still confused about how to use the gradient.
does it mean that on the f '(x) graph if its below the x axis its negative so the f (x) should be decreasing? how do you know what the gradient is? especially for the third image?

Yeah that's it! So if \(f'(x)\) is below the x-axis (is negative), then \(f(x)\) has a negative gradient in that region and is thus decreasing!

Third image? I've only got the two, perhaps missing an attachment?

[Arisa_90]

Does it matter where you draw it decreasing? Like above or below the x axis?
I meant the second diagram in the second image

Yeah that's it! So if \(f'(x)\) is below the x-axis (is negative), then \(f(x)\) has a negative gradient in that region and is thus decreasing!

Third image? I've only got the two, perhaps missing an attachment?

[jamonwindeyer]

Does it matter where you draw it decreasing? Like above or below the x axis?
I meant the second diagram in the second image

Nope! There is no way to know whether it will be above or below - When you draw \(f(x)\) from \(f'(x)\), the axes mean nothing. You don't have the information to know where the intercepts are, because integration introduces an unknown constant.

Right! So in that image, moving from left to right, the gradient starts almost at zero (it is close to flat, but not quiiite flat, so it is a very small negative gradient). It then decreases, becoming more and more negative, approaching a vertical line. So, on the LHS of the graph, the gradient moves from 0 to \(-\infty\). So you just draw any line that moves from 0 to \(-\infty\), never touching either (that's the line in pencil).

On the RHS, it is reversed. We go from a huge positive gradient and slowly approach zero, never touching. This explains the shade of the other side of the graph.

Note that any graph that looks like it does in pencil, shifted UP or DOWN without changing the shape, is also correct.

I think you've got it pretty much nailed by the way you are talking!

[Arisa_90]

I see. I just wanted to ask well if there is an inflection in a f (x) graph does is simply mean it is a either a max or a min point depending on the gradient on either side for the f ' (x) graph? eg. if on the f(x) graph it is decreasing before the inflection point then increasing would it mean that on the f '(x) graph the line will be below the x axis touches the x axis(x intercept where the inflection point was) then goes above x axis?

Nope! There is no way to know whether it will be above or below - When you draw \(f(x)\) from \(f'(x)\), the axes mean nothing. You don't have the information to know where the intercepts are, because integration introduces an unknown constant.

Right! So in that image, moving from left to right, the gradient starts almost at zero (it is close to flat, but not quiiite flat, so it is a very small negative gradient). It then decreases, becoming more and more negative, approaching a vertical line. So, on the LHS of the graph, the gradient moves from 0 to \(-\infty\). So you just draw any line that moves from 0 to \(-\infty\), never touching either (that's the line in pencil).

On the RHS, it is reversed. We go from a huge positive gradient and slowly approach zero, never touching. This explains the shade of the other side of the graph.

Note that any graph that looks like it does in pencil, shifted UP or DOWN without changing the shape, is also correct.

I think you've got it pretty much nailed by the way you are talking!

[jamonwindeyer]

I see. I just wanted to ask well if there is an inflection in a f (x) graph does is simply mean it is a either a max or a min point depending on the gradient on either side for the f ' (x) graph? eg. if on the f(x) graph it is decreasing before the inflection point then increasing would it mean that on the f '(x) graph the line will be below the x axis touches the x axis(x intercept where the inflection point was) then goes above x axis?

Careful, you started correctly by saying that an inflexion point becomes a max or a min. Then you swapped to saying it is an intercept - Inflexions become critical points (maxima/minima!)

So, if your curve is concave down before the inflexion, and concave up after the inflexion, it becomes a maximum. Otherwise, it becomes a minimum

[Arisa_90]

is maxima and minima simply maximum and minium points?
Would the maximum point touch the x axis? Example below.
Must all max and min points touch the x axis?

Careful, you started correctly by saying that an inflexion point becomes a max or a min. Then you swapped to saying it is an intercept - Inflexions become critical points (maxima/minima!)

So, if your curve is concave down before the inflexion, and concave up after the inflexion, it becomes a maximum. Otherwise, it becomes a minimum

[jamonwindeyer]

is maxima and minima simply maximum and minium points?
Would the maximum point touch the x axis? Example below.
Must all max and min points touch the x axis?

Yes they are!

Yes, maxima/minima on \(f(x)\) become x-intercepts on \(f'(x)\)

[Arisa_90]

does that mean all inflections on f(x) will be a max or min that touches the x axis?

Yes they are!

Yes, maxima/minima on \(f(x)\) become x-intercepts on \(f'(x)\)

[Shadowxo]

does that mean all inflections on f(x) will be a max or min that touches the x axis?

There are two types of inflections:
Where f'(x)=0 - stationary point of inflection. For this, f'(x) = zero and f''(x)=0, so it'll be a max/min that touches the x axis on the graph y=f'(x).

Where f'(x)≠0 - non-stationary point of inflection. For this, f'(x)≠0 and f''(x)=0, so it'll be max/min that does not touch the x axis on the graph y=f'(x)

[Arisa_90]

how do you know from the f '(x) graph? How do you know from the attachement above for f'(x) the last major point should touch the x axis?
Also does the y intercept play any important part in the process of graphing f (x)?

There are two types of inflections:
Where f'(x)=0 - stationary point of inflection. For this, f'(x) = zero and f''(x)=0, so it'll be a max/min that touches the x axis on the graph y=f'(x).

Where f'(x)≠0 - non-stationary point of inflection. For this, f'(x)≠0 and f''(x)=0, so it'll be max/min that does not touch the x axis on the graph y=f'(x)

[laurenf58]

Could I have some help with this question please? Thanks!

[kiwiberry]

Could I have some help with this question please? Thanks!

This is in the same form as a + √b, so equating, we get a=6 and b=12!

[jamonwindeyer]

how do you know from the f '(x) graph? How do you know from the attachement above for f'(x) the last major point should touch the x axis?
Also does the y intercept play any important part in the process of graphing f (x)?

Because it is a horizontal point of inflexion, which you know because it flattens out then keeps going again! Think the shape of \(x^3\)!

Nope, y-intercept is pretty useless in these sorts of problems!

[Mathew587]

Hiya,
help with these please

[jamonwindeyer]

Hiya,
help with these please

Hey! For the first one to have real and rational roots, we just need to prove that the discriminant (\\Delta\) is greater than or equal to zero, and that it is a perfect square.



Clearly, in that factorised form, this will never be a negative number (squared numbers are always positive). So, since the discriminant is never negative, there are always real roots. Since it is a perfect square, they are always rational

For that second one, consider the general quadratic \(y=ax^2+bx+c\), and substitute the three pairs of coordinates in to get three equations in terms of \(a\), \(b\) and \(c\) - Solve those simultaneously if you can (otherwise post where you get stuck and we can guide you the rest of the way!)