Question

find the derivative of
$3 {x}^{4} - 8 {x}^{3} + 12 {x}^{2} - 48x + 25$
​

Answer 1

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Q:-find the derivative of

$3 {x}^{4} - 8 {x}^{3} + 12 {x}^{2} - 48x + 25$

$\bf\huge\underbrace{answer}$

------>>>>Here is your answer<<<<--------

$let \: f(x) = 3 {x}^{4} - 8 {x}^{3} + 12 {x}^{2} - 48x + 25$

$➠f'(x) = \frac{d(3 {x}^{4} - 8 {x}^{3} + 12 {x}^{2} - 48x + 25) }{dx}$

$➠f'(x) = 3 \times 4 {x}^{3} - 8 \times 3 {x}^{2} + 12 \times 2x - 48 + 0$

$➠f'(x) = 12 {x}^{3} - 24 {x}^{2} + 24x - 48$

$➠f'(x) = 12( {x}^{3} - 2 {x}^{2} + 2x - 4)$

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Answer 2

Answer:

f(x)=2x³-24x+107 x ∈ [1, 3]

f'(x) =

6 {x}^{2} - 246x

2

−24

To find the points equate f'(x) = 0

\begin{gathered}6 {x}^{2} - 24 = 0 \\ \\ 6( {x}^{2} - 4) = 0 \\ \\ {x}^{2} = 4 \\ x = + 2 \\ x = - 2\end{gathered}

6x

2

−24=0

6(x

2

−4)=0

x

2

=4

x=+2

x=−2

in closed interval x= -2 doesn't lies,so discard x = -2

now find the value of function at x = 1, 2 ,3

f(1) =2(1)³-24(1)+107

= 2-24+107 = 85

f(2) =2(2)³-24(2)+107

= 16-48+107 = 75

f(3) =2(3)³-24(3)+107

= 54-72+107 = 89

So, the function has maximum value in close interval at x= 3, Maximum value= 89.

minimum value at x= 2, minimum value = 75

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