Question

!!!!!!!!!!!!!!!!!!!!!!​

Answer 1

Answer:

don't mowing this question

Answer 2

Please mark me brainliest and follow me

Now, x= $\frac{t^3}{3}$ - $\frac{5t^2}{2}$ + 6t + 1

differentiating x with respect to t,

$\frac{dx}{dt}$ = t² - 5t + 6

as $\frac{dx}{dt}$ =0

$\frac{dx}{dt}$ = t² - 3t - 2t +6 = 0

$\frac{dx}{dt}$ = t(t-3) -2(t-3)

$\frac{dx}{dt}$ = (t-2)(t-3)

t = 2, 3

now, again differentiating,

$\frac{d^2x}{dt^2}$ = 2t -5

if we substitute t= 2,

2(2) - 5 = 4-5= -1

if we substitute t = 3

2(3) - 5 = 6 - 5 = 1

So, the answer is ±1