Question

Consider the function f: R rightarrow R, f(x)=e^x(x^2-a)

Answer 1

Consider the function
\[R \rightarrow R, f(x)=e^x(x^2-a)\]

and \[f'(x)=e^x(x^2+2x-a)\]

Find a value of 'a' that would enable the bracket to be factorised and hence, find the exact coordinates of the stationary points of the graph f(x).

Can you please explain the above question

Answer 2

Let's work this question one part at a time.

`Find a value of 'a' that would enable the bracket to be factorised`

Answer 3

They want you to find a value of `a` that makes the quadratic `x^2+2x-a` factorable.

Answer 4

How about \(a=3\) ?

Answer 5

Can you factor `x^2 + 2x - 3` ?

Answer 6

Yes. (x+3)(x-1) ...

Answer 7

That means a = 3 is good enough, let's look at the next part

Answer 8

` find the exact coordinates of the stationary points of the graph f(x).`

What do you know about the relationship between `stationary points` and the `first derivative f'(x)` ?

Answer 9

if we make f'(x)=0 we will be able to find the stationary points.
So can i write
\[f'(x)=e^x(x^2+2x-3e^x)\]

Answer 10

\[f'(x)=e^x(x^2+2x-3)\]

right ?

Answer 11

yes and that gives us x=-3 and x-1 ..We can go ahead and sub that back into f(x) to find y-coordinate.

So .. any number could have done the trick? like 8 or 24?? is this question a little vague?

Answer 12

The question is not vague as they have just asked you to find ANY ONE value for `a`.

Answer 13

okay. Thank you very much :D I really appreciate your time

Answer 14

It doesn't matter what you pick for `a`, all it requires is the quadratic has to be factorable. They gave you the freedom to choose `a`. This doesn't make the question vague..

Answer 15

Great. Appreciate it!