First Derivatives/ Calculus

Question

Graph f(x)= 1/4x⁴ + 1/3x³ -6x² +2 using first derivative analysis

Answer 1

It helps to find where the function's derivative is 0:

f'(x) = x³+x²-12x

0 = x³+x²-12x

0 = x(x²+x-12)

0 = x(x+4)(x-3)

Thus, x=0, x=3, and x=-4 are the points at which the derivative of f(x) is 0. These correspond to the points (0,2), (3,-81/4), and (-4,-153/4), so these can be graphed.

Hope this helped!

Answer 2

Find f'(x):

(Use the power rule)

f'(x) = x³ + x² - 12x

Find all critical points of f(x)

Critical points are where the derivative equals 0 or the derivative is undefined

f'(x) = 0:

0 = x³+ x² - 12x

0 = x (x² + x - 12)

0 = x (x + 4)(x - 3)

x = 0, -4, 3 are critical points.

f'(x) is defined for all values of x so there are no additional critical points.

Analyze intervals of increase or decrease

Interval Test point f'(x) value f(x) increasing or decreasing

(-∞, -4) x = -5 f'(-5) = -40 decreasing

(-4, 0) x = -1 f'(-1) = 12 increasing

(0, 3) x = 1 f'(1) = -10 decreasing

(3, ∞) x = 5 f'(5) = 90 increasing

For x=-4: f(x) is decreasing before and increasing after so x = -4 is a relative min

For x = 0: f(x) is increasing before and decreasing after so x = 0 is a relative max

For x = 3: f(x) is decreasing before and increasing after so x = 3 is a relative min.

Plug critical points into f(x) and plot the points

f(-4) = 42

f(0) = 2

f(3) = -22.75