Finding the gradient and concavity.

[Eumwilkinson]

Hi completely lost on this question. I don’t understand how to find the gradient ai), even though it’s probably really easy. I have tried sketching the graph of f but it does not cross the point P. Don’t know what I’m doing wrong. I understand how to do aii) but also don’t quite understand b) would really appreciate some help!

 

[MarkFL]

From the graph, and the problem statement we see:

[MATH]f'(4)=-2[/MATH]
This directly answers part (i) of question (a).

What must be the slope of the line normal to \(f\) at \(P\)? Once you know the slope, and given the point, can you find the equation of the requested normal line?

 

[Eumwilkinson]

Hi, thank you. Still struggling to understand how you got to -2 - where did you get the 4 from? Point P? Sorry, I really struggle to get my head around this!

 

[MarkFL]

Yes, we are given that the point \((4,-2)\) lies on the curve of \(f'(x)\). This directly tells us that:

[MATH]f'(4)=-2[/MATH]

 

[Eumwilkinson]

So how is point P relevant in the question? Sorry for so many questions, just this really confuses me.

 

[MarkFL]

Point \(P\) will allow you to find the normal line to \(f\) and passing through \(P\). To find the equation of a line, we need both the slope and a point on the line. The slope will come from the negative multiplicative inverse of the gradient of \(f\) at \(P\) and the point is \(P\) itself.

 

[pka]

So how is point P relevant in the question? Sorry for so many questions, just this really confuses me.

Suppose that \(\displaystyle Q: (c.d)\) on the graph of the function \(\displaystyle f\). If \(\displaystyle f'(c)=s\ne 0\) then the equation of the normal to the curve of \(\displaystyle f\) through the point \(\displaystyle Q\) is \(\displaystyle y=\frac{-1}{s}(x-c)+d\).

 

[Steven G]

The derivative, f'(x), is the slope formula for f(x). What exactly does that mean? Lets look at an example.

Suppose f(x) = x². Then f'(x) =2x. This f'(x) function gives the slope of the tangent for f(x).
Suppose on f(x) = x² we want to know the slope of the tangent line at the point (3, 9). We just compute f'(3) = 2*3=6
If we wanted to know the slope of the tangent for f(x) at (-2, 4), the we compute f'(-2) = -4. So the slope of the tangent kine at (-2,4) is -4

Now back to your question. You are given the point (4,-2) on f'(x). So the slope of the tangent line of f(x) at (4,3) is f'(4)=-2. Note that both points just mentioned have the same x-values. Do you understand what that means.

Can you continue from here? If not, then please let us know where you are confused.