Shayan A.

asked • 04/10/17

Find the set of values of k for which f(x)= 3x^2 -5x -k is greater than 1 for all real x

This is a year 11 methods question about applying function notation.
Could someones please explain the question and then give me a working out for it?

Prentice D.

Ok so this question is asking for what values for k such that the graph never goes below 1. Since this is a hyperbolic function of the form f(x)=ax2 +bx+c (happy face graph due to positive a value), the function will have a lowest point when the derivative of this function is equal to zero. So,
f(x) = 3x2 -5x -k
f'(x) = 6x -5   (notice that the derivative does not depend on k, so the lowest point of the graph will have the same x value no matter the k value).
now sub in f'(x) = 0
0 = 6x-5
x=5/6
 
Now from the original function sub in f(x) = 1, and x=5/6
1 = 3*(5/6)^2 -5(5/6) -k
k=-37/12
k=-3.08 (approximately)
Now the k value will only lower or increase the height of the graph, so increasing the value for k value will put the lowest part on the graph above 1, while decreasing 1 will make the lowest part on the graph less than 1.
So therefore the value for k must be k>-3.08, so that the graph is greater than 1 for all real x.
 
 
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04/10/17

Shayan A.

Thank you for your response,
It makes sense using the derivative (I had some personal practice on it before) but we havent officially been taught yet. So I was wondering if there is another way of solving this question.
 
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04/10/17

1 Expert Answer

By:

Roman C. answered • 04/10/17

Masters of Education Graduate with Mathematics Expertise

Shayan A.

Thank you for your response,
I was just wondering where does the y=-(discriminant)/4a comes from?
Whats the reason that it works?
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04/10/17

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