Anyone know how to solve this?

[tough bridge]

is it 0.3125

[hidden depot]

a•x^b, derivative is (ab)•x^(b-1)

Do this to find full derivative, then find the zero of that derivative to find maximum

[tough bridge]

ty

[tough bridge]

do you know if this is right

[hidden depot]

That is correct. Find the derivative

[tough bridge]

is the derivative y' = -16x + 5

and then substitute 0.3125?

[hidden depot]

Yes

[tough bridge]

it gives me 0 tho

[hidden depot]

Yes
Given a function f(x), if the derivative of f (usually notated as f’(x))equals zero at some x, then there is a local minimum or maximum of f(x) at x

Since .3125 gives zero on f’(x), .3125 is also where the local maximum is for f(x)

[tough bridge]

thank you

[high estuary]

note that for quadratics you can instantly tell whether the extremum is max or min by the coefficient of the x² (positive and it's the min, negative and it's the max) @tough bridge

[tough bridge]

cool, thank you

also

is it D?

[high estuary]

that's a reflection in the y axis

because if you reflect something along the y axis it's y coordinate will not change, it's the X coordinate that becomes negative

@tough bridge

[tough bridge]

So b yeh

[high estuary]

yes

[tough bridge]

thank u

[high estuary]

kinda neat that that quadrilateral is a square

[tough bridge]

is this 41?

[high estuary]

yes

[tough bridge]

thanks

[tough bridge]

You sure ?

it's 41

[high estuary]

16x+9
16(2)+9
32+9
41

certain

[tough bridge]

thank u

is this one B?

[high estuary]

split it and take the constant out
2∫(1/X)dx + ∫3sin(3x)dx
3x=u
3dx=du
du/3 = dx
∫sin(u)dx
2ln(x)+(-cos(3x))
2ln(x)-cos(3x)+C

[tough bridge]

perfect, thank you so much

life saver u are

[high estuary]

np

[tough bridge]

is this one D?

[high estuary]

weird. none of those answers are right

[tough bridge]

Yeh

i got 14.62

and 8.19

so im not sure

[high estuary]

wait no

it's d yeah

I made an error in my sd

[tough bridge]

Ah

aight

ty

[tough bridge]

Idk why i got 14.62