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Find the absolute maximum and minimum values of the function over the indicated​ interval, and indicate the​ x-values at which they occur. f (x )equalsx Superscript 4 Baseline minus 18 x squared plus 9​; [negative 5 comma 5 ]

User Kralyk
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1 Answer

2 votes

Answer:

absolute maximum = 184

absolute minimum = -72

Explanation:

Given the function f(x) = x⁴-18x²+9 at the interval [-5, 5], the absolute maximum and minimum values at this end points are as calculated;

at end point x = -5

f(-5) = (-5)⁴-18(-5)²+9

f(-5) = 625-450+9

f(-5) = 184

at end point x = 5

f(5) = (5)⁴-18(5)²+9

f(5) = 625-450+9

f(5) = 184

To get the critical point, this points occurs at the turning point i.e at

dy/dx = 0

if y = x⁴-18x²+9

dy/dx = 4x³-36x = 0

4x³-36x = 0

4x (x²-9) = 0

4x = 0

x = 0

x²-9 = 0

x² = 9

x = ±3

Using the critical points [0, ±3]

when x = 0, f(0) = 0⁴-18(0)+9

f(0) = 9

Similarly when x = 3, f(±3)= (±3)⁴-18(±3)²+9

f(±3) = 81-162+9

f(±3) = -72

It can be seen that the absolute minimum occurs at x= ±5 and and absolute minimum occurs at x =±3

absolute maximum = 184

absolute minimum = -72

User Florin Vistig
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5.7k points