Sketch a function with f'(x)0 for x

Question

Sketch a function with f'(x)>0 for x<2, f'(2)=0 and f'(x)<0 when x>2

campbell_st · Answer

so there is a stationary point at x = 2

the slope of the tangent to the curve is positive for x < 2 

and the slope of the tangent is negative for x > 2

I hope this helps...

anonymous · Answer

Yes I know that but still having troubles with it, please help?

campbell_st · Answer

ok... well to me, based on the information, its a concave down parabola with a vertex at x = 2 any you have no information about f(2)

campbell_st · Answer

I hope this helps you... the curve may be negative definite, touch the axis at x = 2 or have 2 unequal real roots

anonymous · Answer

Could you please tell me how you know the vertex is 2, and that it is a concave down?

campbell_st · Answer

well the 1st derivative.... 

stationary points in any curve occur when the 1st derivative is equal to zero and solved... 


e.g. f(x) = x^2 + 6x

f'(x) = 2x + 6 

let f'(x) = 0

then    2x + 6 = 0 

so x = -2 is a stationary point... or has a slope of zero at x = -3

so looking at the slope of the tangents in your question
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anonymous · Answer

Ohhh ok I see it now

anonymous · Answer

Thank You for your time