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KIM [24]
3 years ago
12

URGENT O_O Find the roots of the polynomial equation! 2x^3+2x^2-19x+20=0

Mathematics
1 answer:
ohaa [14]3 years ago
4 0

The function is

  • y=2x^3+2x^2-19x+20

Graph attached

There is one real zero only

That's-1.047

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Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separ
Sidana [21]

Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

f(x)=(x-3)^{-2}

When we differentiate this function with respect to x, we get;

f'(x)=-\frac{2}{(x-3)^3}

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

0.06-0.25=-\frac{2}{(c-3)^3} (6)

-0.19=-\frac{12}{(c-3)^3}

(c-3)^3=\frac{-12}{-0.19}

(c-3)^3=63.15789

c-3=\sqrt[3]{63.15789}

c=3+\sqrt[3]{63.15789}

c=6.98

If this function satisfies the Mean Value Theorem, then f must be continuous on  [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.

 

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