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Find all relative extrema of the function. (If an answer does not exist, enter DNE.)f(x) = x4 − 32x + 8relative maximum (x, y) = relative minimum (x, y) =

Question

Find all relative extrema of the function. (If an answer does not exist, enter DNE.)
f(x) = x4 − 32x + 8
relative maximum (x, y) =

 


relative minimum (x, y) =

 

check_circleAnswer
Step 1

Given,

f(x) x4-32x + 8 = 0
Let's find the critical point of the above function. For finding the critical point,
we find all the x for which f'(x) = 0.
d(x")
using
dC
f'(x) 4x3 32
пх^-1
dx
dx
For critical point, f'(x) = 0
=4x3 32 = 0
=> x = 2
help_outline

Image Transcriptionclose

f(x) x4-32x + 8 = 0 Let's find the critical point of the above function. For finding the critical point, we find all the x for which f'(x) = 0. d(x") using dC f'(x) 4x3 32 пх^-1 dx dx For critical point, f'(x) = 0 =4x3 32 = 0 => x = 2

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Step 2

Also,

...
Now we will use the double derivative test to check whether x = 2 is
relative maxima or minima
d(x")
using
f"(x) = 12x2
пх*-
dx
f"(2) 12(2)2 = 48
As f"(2)>0 => x= 2 is a point of relative minima
Аnd the miniтит value — f (2)
24-32(2)8
- 40
Also, relative maxima of f (x) does not exist
help_outline

Image Transcriptionclose

Now we will use the double derivative test to check whether x = 2 is relative maxima or minima d(x") using f"(x) = 12x2 пх*- dx f"(2) 12(2)2 = 48 As f"(2)>0 => x= 2 is a point of relative minima Аnd the miniтит value — f (2) 24-32(2)8 - 40 Also, relative maxima of f (x) does not exist

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Math

Calculus

Derivative

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