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TEA [102]
3 years ago
13

Consider the following equation. cos x = x3 (a) Prove that the equation has at least one real root. f(x) = cos x − x3 is continu

ous on the interval [0, 1], f(0) = 0, and f(1) = cos 1 − 1 ≈ −0.46 0. Since 0 −0.46, there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x − x3 = , or cos x = x3, in the interval (0, 1). (b) Use your calculator to find an interval of length 0.01 that contains a root.

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0

Answer:

b. 0.86, 0.87

Step-by-step explanation:

a. Find attached solution to a

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Answer:

The given point A (6,13) lies on the equation.  True

The given point  B(21,33) lies on the equation.  True

The given point  C (99, 137) lies on the equation.  True

Step-by-step explanation:

Here, the given equation is : y =\frac{4}{3} x + 5

Now,check the given equation for the given points.

1)  A (6,13)

Substitute x = 6 in the given equationy =\frac{4}{3} x + 5

y =\frac{4}{3} (6) + 5  = 4(2) + 5  = 8 + 5 =  13

⇒ y= 13

Hence, the given point A (6,13) lies on the equation.

2)  B  (21,33)

Substitute x = 21 in the given equationy =\frac{4}{3} x + 5

y =\frac{4}{3} (21) + 5  = 4(7) + 5  = 28 + 5 =  33

⇒ y  = 33

Hence, the given point  B(21,33) lies on the equation.

3)  C (99, 137)

Substitute x = 99 in the given equationy =\frac{4}{3} x + 5

y =\frac{4}{3} (99) + 5  = 4(33) + 5  = 132 + 5 =  137

⇒ y  = 137

Hence, the given point  C (99, 137) lies on the equation.

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