Find the values of x, such that f(x)= 2x³ - 15x² + 36x + 1 is an increasing function.

238.9608076923

Step-by-step explanation

To get the answer first we have to divide 15.361 by 26 and nex add to it 233+5.37

answer:

35/36

Step-by-step explanation:

this is the correct answer...

answer:

36/15=6/5

₹6/5 answer

Step-by-step explanation:

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Explanation:

the maximum value=29

the minimum value=28

f(x) = 2x3 - 15x2 + 36x + 1

f'(x) = 6x2 - 30x + 36

= − 6(x2 - 5x + 6)

= 6(x - 2)(x - 3)

f(x) is increasing in [0, 2] and decreasing in [2, 3]. f (x) is many one.

f(0) = 1 f(2)

= 29 f(3)

= 28 Range is [1, 29].

Hence, f (x) is many-one-onto.

f(x) = 2x3 - 15x2 + 36x + 1

f'(x) = 6x2 - 30x + 36

= − 6(x2 - 5x + 6)

= 6(x - 2)(x - 3)

f(x) is increasing in [0, 2] and decreasing in [2, 3]

. f (x) is many one.

f(0) = 1

f(2) = 29

f(3) = 28

Range is [1, 29].

Hence, f (x) is many-one-onto.

Hope this will help you

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maxima will be 29 .

minima will be 28.

Step-by-step explanation:

ANSWER

f(x)=2x

3

−15x

2

+36x+1,f:[0,3]→[1,29]

1. Check for one-one mapping-

f(x) is a cubic polymonial. First derivative of f(x) gives f

′

(x)=6x

2

−30x+36.

Solve f

′

(x)=0 to get the critical points.

x

2

−5x+6=0

(x−3)(x−2)=0

⇒x∈{3,2}

f(x) is increasing for x<2 or x>3 (∵f

′

(x)>0) and f(x) is decreasing for 2<x<3 (∵f

′

(x)<0).

Hence, f(x) is not strictly increasing or strictly decreasing in the entire domain. So f(x) is not one-one.

2. Check for onto mapping-

f(x) has maximum at x=2

f(2)=29, which is maximum value of f.

(∵ in the given domain [0,3], f is increasing in [0,2) and decreasing in (2,3].)

f(0)=1 and f(3)=28

Hence, range of f(x)=[1,29] which is equal to the co-domain. Hence, f(x) is onto.

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answer:

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