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How do you find all critical point and determine the min, max and inflection given #f(x)=x^2-8x-10#?

Answer 1

At #x=4# is a minimum only

#f(x)=x^2-8x-10# Differentiating with respect to x, #f'(x)=2x-8# Differentiating once more, #f''(x)=2# which is #>0# so we are expecting a minimum When #f'(x)=0 => 2x-8=0 =>x=4# When #x<4, f'(x) <0# and when #x>4, f'(x)>0# So we have a minimum Check #f(3)=3*3-8*3-10=-25# #f(5)=5*5-8*5-10 =-25# and #f(4)=4*4-8*4-10=-26# So we have a min. only

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Answer 2

Emily Ammerman

To find critical points, set the derivative of ( f(x) ) equal to zero and solve for ( x ). Then, classify each critical point using the second derivative test. ( f'(x) = 2x - 8 ). Critical point: ( x = 4 ). ( f''(x) = 2 ). Since ( f''(4) > 0 ), it's a local minimum. No inflection points.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.