What is the extreme (x^2-15)/(x-4) ?

The extreme (x^2-15)/(x-4) is Maximum(3,6),Minimum(5,10)

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extreme (x^2-15)/(x-4)

extreme points

\frac{x ^{2} - 15}{x - 4}

Maximum (3, 6), Minimum (5, 10)

Solution steps

f^{'} (x) = \frac{x ^{2} - 8 x + 15}{( x - 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < 4,

Decreasing

: 4 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 3

into

\frac{x ^{2} - 15}{x - 4} : 6

Maximum (3, 6)

Plug

x = 5

into

\frac{x ^{2} - 15}{x - 4} : 10

Minimum (5, 10)

Maximum (3, 6), Minimum (5, 10)

f (x) = e^{x}, 0 \leq x \leq ln (18)

f (x) = 2 x^{3} - 6 x^{2} - 210 x + 7

f (x) = - 3 x^{5} + 5 x^{3} + 1

f (x) = - 2 + x - x^{2}

f (x, y) = x^{2} + y^{2} + 4 x - 2 y + 6

What is the extreme (x^2-15)/(x-4) ?
The extreme (x^2-15)/(x-4) is Maximum(3,6),Minimum(5,10)