Question

The image of the interval [1,3] under the
mapping f:R→R, given by
f(x) = 2x” – 24x+107
is

Answer 1

Answer:

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Answer 2

SOLUTION

TO DETERMINE

The image of the interval [1,3] under the mapping f : R → R given by f(x) = 2x³ - 24x + 107

EVALUATION

Here the mapping f : R → R given by

f(x) = 2x³ - 24x + 107

So f(x) is a polynomial function

So it is continuous for all real values of x

So f(x) is continuous in interval [ 1 , 3 ]

Now

$\sf{f(1) }$

$\sf{ = 2 \times {(1)}^{3} - (24 \times 1) + 107 }$

$\sf{ = 2 - 24 + 107 }$

$\sf{= 85 }$

$\sf{f(2) }$

$\sf{= 2 \times {(2)}^{3} - (24 \times 2) + 107 }$

$\sf{ = 16 - 48 + 107 }$

$= \sf{75}$

$\sf{f(3) }$

$\sf{= 2 \times {(3)}^{3} - (24 \times 3) + 107 }$

$\sf{ = 54 - 72 + 107 }$

$\sf{ = 89}$

So the minimum value of f(x) = 75 at x = 2

Maximum value of f(x) = 89 at x = 3

Hence the required image of the interval [1,3] under the mapping f : R → R given by f(x) = 2x³ - 24x + 107 is [ 75 , 89 ]

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