D. B Arr -SA 4 -3 (a) Is there a value x, for -1 ≤ x ≤ 4, such that g(x) = 0? Justify your answer. of 4 + √²₁ 8(+)dt = 4 + [0=0] = 4 4 f(x) The figure above shows the graph of the continuous function f. The regions A, B, C, and D have areas 4, 13, 16, and 3, respectively. For -6 ≤ x ≤ 6, the function g is defined by g(x) = 4 + f2₁f(t) dt. (b) Find the absolute minimum value of g on the interval -6 ≤ x ≤ 6. g'changes sign (- > +) g₁ = g(t) g(x) = f (c) Find the value of ¹ f(5-x) dx 11 there is no value of x for glx) = C absolute minimum at x = 1. since f(x) changes sign (-)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
6.
y
f(x)
B
13
Ar
+
-SA 4 -3 -2
2
4
А
(a) Is there a value x, for -1 ≤ x ≤ 4, such that g(x) = 0? Justify your answer.
4
4 + 1 ² ₁ 8(+)dt = 4+ [0-0] =4
7
cx
The figure above shows the graph of the continuous function f. The regions A, B, C, and D have areas 4, 13,
16, and 3, respectively. For -6 ≤ x ≤ 6, the function g is defined by g(x) = 4 + ₁ f(t) dt.
(b) Find the absolute minimum value of g on the interval -6 ≤x≤ 6.
g' changes sign (--> +)
g = g(t)
-1
(c) Find the value of ff(5-x) dx
net Dron
g(x) = {
g(x) = ff
there is no value of x for g(x)=0
absolute minimum at x = 1
since
f (x) changes sign (-) → (+).
St. Patrick's Day
√₁ 8(5-x) = g(
s(5+1) - g(5-1) = g(6)-g(4) =
Transcribed Image Text:6. y f(x) B 13 Ar + -SA 4 -3 -2 2 4 А (a) Is there a value x, for -1 ≤ x ≤ 4, such that g(x) = 0? Justify your answer. 4 4 + 1 ² ₁ 8(+)dt = 4+ [0-0] =4 7 cx The figure above shows the graph of the continuous function f. The regions A, B, C, and D have areas 4, 13, 16, and 3, respectively. For -6 ≤ x ≤ 6, the function g is defined by g(x) = 4 + ₁ f(t) dt. (b) Find the absolute minimum value of g on the interval -6 ≤x≤ 6. g' changes sign (--> +) g = g(t) -1 (c) Find the value of ff(5-x) dx net Dron g(x) = { g(x) = ff there is no value of x for g(x)=0 absolute minimum at x = 1 since f (x) changes sign (-) → (+). St. Patrick's Day √₁ 8(5-x) = g( s(5+1) - g(5-1) = g(6)-g(4) =
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