Select the correct comparison Set A Set B?
2 answers:
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(a) It looks like you're saying
Find the critical points of F(x). By the fundamental theorem of calculus,
F'(x) = x - 3x² + 7
The critical points are where the derivative vanishes. Using the quadratic formula,
x - 3x² + 7 = 0 ⇒ x = (1 ± √85)/6
Compute the second derivative of F :
F''(x) = 1 - 6x
Check the sign of the second derivative at each critical point.
• x = (1 + √85)/6 ≈ 1.703 ⇒ F''(x) < 0
• x = (1 - √85)/6 ≈ -1.370 ⇒ F''(x) > 0
This tells us F attains a minimum of
(b) Split up the domain of F at the critical points, and check the sign of F'(x) over each subinterval.
• over (-∞, -1.370), consider x = -2; then F'(x) = -7 < 0
• over (-1.370, 1.703), consider x = 0; then F'(x) = 7 > 0
• over (1.703, ∞), consider x = 2; then F'(x) = -3 < 0
This tells us that
• F(x) is increasing over ((1 - √85)/6, (1 + √85)/6)
• F(x) is decreasing over (-∞, (1 - √85)/6) and ((1 + √85)/6, ∞)
(c) Solve F''(x) = 0 to find the possible inflection points of F(x) :
F''(x) = 1 - 6x = 0 ⇒ 6x = 1 ⇒ x = 1/6
Split up the domain at the inflection point and check the sign of F''(x) over each subinterval.
• over (-∞, 1/6), consider x = 0; then F''(x) = 1 > 0
• over (1/6, ∞), consider x = 2; then F''(x) = -11 < 0
This tells us that
• F(x) is concave up over (-∞, 1/6)
• F(x) is concave down over (1/6, ∞)
The excluded values are x = 4 and x - 1
<h3>How to determine the excluded values?</h3>The complete question is added as an attachment
The function is given as:
(2x^2 - 7x - 4)/(x^2 - 5x + 4)
Set the denominator to 0
x^2 - 5x + 4 = 0
Expand
x^2 - x - 4x + 4 = 0
Factorize the equation
x(x -1) - 4(x - 1) = 0
This gives
(x - 4)(x - 1) = 0
Solve for x
x = 4 and x - 1
Hence, the excluded values are x = 4 and x - 1
Read more about excluded values at
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