# AS questionWatch

#1
CIE w11qp11
4 A function f is defined for x ∈ R and is such that f′(x) = 2x − 6. The range of the function is given byf(x) ≥ −4.(i) State the value of x for which f(x) has a stationary value. [1](ii) Find an expression for f(x) in terms of x.

I don't understand ii):
MS says to substitute 3, -4 into the integrated f(x) equation, why?
0
2 years ago
#2
(Original post by Deathlotus)
CIE w11qp11
4 A function f is defined for x ∈ R and is such that f′(x) = 2x − 6. The range of the function is given byf(x) ≥ −4.(i) State the value of x for which f(x) has a stationary value. [1](ii) Find an expression for f(x) in terms of x.

I don't understand ii):
MS says to substitute 3, -4 into the integrated f(x) equation, why?
If f(x) is a quadratic and the range is f(x) ≥ −4 then it must mean that the minimum point is at y = -4.

Does that make sense? From that you can find the x - coordinate.
0
#3
(Original post by notnek)
If f(x) is a quadratic and the range is f(x) ≥ −4 then it must mean that the minimum point is at y = -4.

Does that make sense? From that you can find the x - coordinate.
Thanks for your reply, but why can't the stationary value be say at -3 or -2?
0
2 years ago
#4
(Original post by Deathlotus)
Thanks for your reply, but why can't the stationary value be say at -3 or -2?
If it was something other than a quadratic then yes there could be multiple stationary poins within this range.

But this is a U shaped parabola which has a single minimum point. Draw a sketch to help you.

The range says f(x) ≥ −4. This means that the curve passes through points with y = -4 and any values bigger than this. So y = -4 has to be the minimum.
0
#5
Thank you! That sketch helped me understand now.
0
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