Maxima and minima help?

Question

I am confused on this question.

Let f be the function given by f(x) = 2 ln(x^2 - 5) - x with the domain of 3 less than or equal to x less than or equal to 6.

First, Find the x-coordinate of relative maximum point of f if exists. Justify this part using calculus.

Second, Find the absolute maximum value of f(x) on 3 less than or equal to x less than or equal to 6.

Finally, write an equation of the line tangent to the graph of f at x=4.

I know it looks a bit long, but once you start, you'll be on a roll.

Thanks!

Answer 1

f '(x) = 2*2x / (x^2 - 5) - 1

Stationary where f '(x) = 0

therefore

4x / (x^2 - 5) - 1 = 0

4x - (x^2 - 5) = 0

Multiply each term by -1:

x^2 - 4x - 5 = 0

(x - 5)(x + 1) = 0

x = 5 or -1

Discard -1 because (-1)^2 -5 < 0 hence ln is not defined; anyway, it's outside the given domain.

Now use the quotient formula to get

f "(x) = [4(x^2 - 5) - 2x*4x] / [(x^2 - 5)^2]

...... = [-4x^2 + 20] / [(x^2 - 5)^2]

...... = - 80 / 400 when x = 5

Therefore x = 5 gives a maximum turning point.

Second:

f(3) = 2 ln 4 - 3

f(5) = 2 ln 20 - 5

... = 2 ln 4 + 2 ln 5 - 5

... = 2 ln 4 - 3 + 2 ln 5 - 2

... > 2 ln 4 - 3 since 2(ln 5 - 1) > 0

f(6) = 2 ln 31  - 6 which we can show is less than f(5)

Hence f(5)  = 2 ln 20 - 5 is the absolute maximum in this domain.

ALTERNATIVELY we can do some algebra shuffling to get

f '(x) = - [x^2 - 4x - 5] / (x^2 - 5)

..... = -[(x - 5)(x + 1)] / (x^2 - 5)

and hence when 3 < x < 5, only x - 5 is negative and so f '(x) > 0;

and when 5 < x < 6, each factor is positive and so f'(x) < 0

This proves that x = 5 gives a local maximum, also since the function is increasing from x = 3 to x =5 and decreasing from x = 5 to x = 6, the absolute maximum is at x = 5.

Finally, f(4) = 2 ln 11 - 4,

and f '(4) = 5/11

hence the equation of the tangent at x = 4 is

y - y1 = m(x - x1)

y - (2 ln 11 - 4) = (5/11)(x - 4)

11y - 22 ln 11 + 44 = 5x - 20

5x - 11y + 22 ln 11 - 64 = 0