Question

x equal to minus 3 x square + 5 x minus 12 has maximum value.

plz make it fast​

Answer 1

Step-by-step explanation:

Answer

Correct option is

D

−

12

1

p(x)=3x

2

−5x+2

At maximum,

dx

d

p(x)=p

′

(x)=0

⟹6x−5=0

⟹x=

6

5

Hence maximum value of p(x) is

p(

6

5

)=3(

6

5

)

2

−5(

6

5

)+2

=

12

−1

Answer-(D)

Answer 2

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x = $-3x^2+5x-12$ has maximum value at $-\dfrac{119}{12}$

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The maximum value of a quadratic function can be calculated by

$c - \dfrac{b^2}{4a}$

So, in the question:

$-3x^2+5x-12$ = $ax^2+bx+c$

where,

a = - 3

b = 5

c = - 12

Substituting the value of a, b and c in the equation,

$f_{max} = -12-\dfrac{25}{-12}\\\\f_{max}=-12+\dfrac{25}{12}\\\\f_{max}=\dfrac{-144+25}{12}\\\\f_{max}=-\dfrac{119}{12}$

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