What is the extreme y=x^3-15x^2+48x+300 ?

The extreme y=x^3-15x^2+48x+300 is Maximum(2,344),Minimum(8,236)

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extreme y=x^3-15x^2+48x+300

extreme points

y = x^{3} - 15 x^{2} + 48 x + 300

Maximum (2, 344), Minimum (8, 236)

Solution steps

f^{'} (x) = 3 x^{2} - 30 x + 48

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 2

into

x^{3} - 15 x^{2} + 48 x + 300 : 344

Maximum (2, 344)

Plug

x = 8

into

x^{3} - 15 x^{2} + 48 x + 300 : 236

Minimum (8, 236)

Maximum (2, 344), Minimum (8, 236)

f (x) = y (x - 5)

2 x^{2} - 16 x + 26

f (x) = Y (x) = x^{2} - 5 x + 6

f (x) = \frac{1}{8} (7 x - 1), x \leq 3

f (x) = \frac{( x ^{2} )}{x ^{2} - 4} + 2

What is the extreme y=x^3-15x^2+48x+300 ?
The extreme y=x^3-15x^2+48x+300 is Maximum(2,344),Minimum(8,236)