VCE Methods Question Thread!

[grannysmith]

Oh, so if the point exceeds the stationary point, then it is positive, right? And if it is under the stationary point, then the gradient is negative!?

Because this is a negative parabola, to the left of the stationary the gradient is positive and to the right of it the gradient is negative. So 'under' the stationary point, you've got both pos and neg gradients.

Think of the difference between a negative and a positive linear graph. Now take two points on either side of the stationary point and draw a chord. You'll see that on the left, the chord is akin to a positive linear graph and to the right, it is similar to negative linear graph.

[cosine]

Because this is a negative parabola, to the left of the stationary the gradient is positive and to the right of it the gradient is negative. So 'under' the stationary point, you've got both pos and neg gradients.

Think of the difference between a negative and a positive linear graph. Now take two points on either side of the stationary point and draw a chord. You'll see that on the left, the chord is akin to a positive linear graph and to the right, it is similar to negative linear graph.

Yep i totally understand that! But, how does that mean the gradient is always decreasing?

[grannysmith]

Yep i totally understand that! But, how does that mean the gradient is always decreasing?

Well I like to think of the derivative function of the negative parabola i.e. a negative linear graph which is always decreasing. Now the 'y-axis' of the derivative function describes the gradient of the negative parabola (i.e. dy/dx). As the y-coordinate of a negative linear graph is always decreasing (as the x-value increases), therefore the gradient of the original function is also always decreasing

But another way of looking at is that as the parabolic function approaches a stationary point (i.e. gradient = 0), the gradient must then be decreasing, say, from 5 to 0. Then after the stationary point, the gradient is actually negative and it continues to decrease as the graph becomes 'steeper'.

[cosine]

Also what does it mean by: (the image attached)


I know that when x=2, f(x)=3. But, whats this got to do with limits? Or am i missing key knowledge from limits that i should visit before doing these questions? Thank you

[pi]

Gonna use random pics from google images that are similar.

Here is what your graph is like:


Looking at the part of the graph to the right of the turning point:


Hence, the gradient is getting MORE negative (ie. is DECREASING) as you go further away to the right from the turning point.

The part of the graph to the left of the turning point starts off very steep (ie. large magnitude positive) and becomes less and less positive (ie. DECREASING) until it hits the turning point (gradient of zero) and then the right side of the graph as above.

Hence, always DECREASING.

Also what does it mean by: (the image attached)


I know that when x=2, f(x)=3. But, whats this got to do with limits? Or am i missing key knowledge from limits that i should visit before doing these questions? Thank you

Well f(2) doesn't exist as the function is discontinuous at that point. Hence for f(x) can only approach 3 (as a limit).

[cosine]

Thank you so much Pi!!

[dankfrank420]

Cosine, little tip if you're not understanding it.

Using a ruler, look at how the tangent of the parabola changes over time.

[cosine]

For the attached, I know that if we substitute in x=0, f(x) would not exist. But, when we graph it, x=0 does exist. So what exactly does limit mean and stuff.. yeah Thanks

[TrueTears]

For the attached, I know that if we substitute in x=0, f(x) would not exist. But, when we graph it, x=0 does exist. So what exactly does limit mean and stuff.. yeah Thanks

What do you mean by "x=0 does exist"?

[Maths Forever]

For the attached, I know that if we substitute in x=0, f(x) would not exist. But, when we graph it, x=0 does exist. So what exactly does limit mean and stuff.. yeah Thanks

You need to take out x as a common factor.

Limit as x approaches 0 of x(x+3) / x = limit as x approaches 0 of (x + 3) = 0 + 3 = 3

So as x approaches 0, f(x) approaches 3.

Sorry for the lack of mathematical symbols. Hope this helps.

[Maths Forever]

For the attached, I know that if we substitute in x=0, f(x) would not exist. But, when we graph it, x=0 does exist. So what exactly does limit mean and stuff.. yeah Thanks

Furthermore, a limit means the value of f(x) that a function will approach as 'x' approaches a given value.

[TrueTears]

You need to take out x as a common factor.

Limit as x approaches 0 of x(x+3) / x = limit as x approaches 0 of (x + 3) = 0 + 3 = 3

So as x approaches 0, f(x) approaches 3.

Sorry for the lack of mathematical symbols. Hope this helps.

But why does ?

[cosine]

Oh alright I get it now, thank you guys!

So the reason why we had to factor out the x was because if we didn't, we would have an undefined result as 0 as the denominator does not exist, right? And limit x------> 3 means to find the limit or f(x) value when x is approaching 3, right?


Thank you for the help!

[Maths Forever]

But why does ?

You substitute whatever value 'x' is approaching to represent 'x' in 'x + 3'. In this case x is approaching 0.

Let's look at the graph y = x. If x approaches 5, then so will y. Since there is the coordinate (5, 5) on a y = x graph.

Hence, 0 + 3 = 3.

[Maths Forever]

Oh alright I get it now, thank you guys!

So the reason why we had to factor out the x was because if we didn't, we would have an undefined result as 0 as the denominator does not exist, right? And limit x------> 3 means to find the limit or f(x) value when x is approaching 3, right?


Thank you for the help!

Yes that is correct!