In exercise, find the relative extrema of the function.
f(x) = x4 - 2x2 + 5

minima at x = ±1

Step-by-step explanation:

Given function in the question:

f(x) = x⁴ - 2x² + 5

now,

To find the extrema differentiating the function with respect to 'x' and equate it to zero

we get

f'(x) = 4x³ - (2)(2x) + 0 = 0

or

4x³ - (2)(2x) = 0

or

4x³ - 4x = 0

or

x³ - x = 0

or

x(x² - 1) = 0

or

x = 0 and x = ±1

checking for maxima or minima

again differentiating f'(x) with respect to x, we get

f''(x) = (3)4x² - x

or

f''(x) = 12x² - x

at x = 0

f"(0) = 12(0)² - 0 = 0          [neither maxima nor minima]

at x = -1

f"(-1) =  12(-1)² - (-1) = 12 + 1 = 13  > 0      [relative minima]

at x = 1

f"(1) =  12(1)² - 1 = 12 - 1 = 11  > 0      [relative minima]

hence,

minima at x = ±1

it should be a and d. lmk if i'm wrong xd

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional