What are the local extrema of #f(x)= x^3 - 9x^2 + 19x - 3 #?

Answer 1

#f(x)_max =(1.37, 8.71)#
#f(x)_min =(4.63, -8.71)#

#f(x)= x^3-9x^2+19x-3#
#f'(x) = 3x^2-18x+19#
#f''(x) = 6x-18#
For local maxima or minima: #f'(x) =0#
Thus: #3x^2-18x+19 =0#

Applying the quadratic formula:

#x=(18+-sqrt(18^2-4xx3xx19))/6#
#x=(18+-sqrt96)/6#
#x=3+-2/3sqrt6#
#x~= 1.367 or 4.633#

To test for local maximum or minimum:

#f''(1.367) < 0 -># Local Maximum
#f''(4.633) > 0 -># Local Minimum
#f(1.367) ~= 8.71# Local Maximum #f(4.633) ~= -8.71# Local Minimum
These local extrema can be seen on the graph of #f(x)# below.

graph{ x^3-9x^2+19x-3 [-22.99, 22.65, -10.94, 11.87]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the local extrema of ( f(x) = x^3 - 9x^2 + 19x - 3 ), we first take the derivative of the function and set it equal to zero to find critical points. Then, we analyze the second derivative to determine the nature of these critical points.

( f'(x) = 3x^2 - 18x + 19 )

Setting ( f'(x) ) equal to zero:

( 3x^2 - 18x + 19 = 0 )

Using the quadratic formula:

( x = \frac{{18 \pm \sqrt{{(-18)^2 - 4(3)(19)}}}}{{2(3)}} )

( x = \frac{{18 \pm \sqrt{{324 - 228}}}}{{6}} )

( x = \frac{{18 \pm \sqrt{{96}}}}{{6}} )

( x = \frac{{18 \pm 4\sqrt{6}}}{{6}} )

( x = \frac{{9 \pm 2\sqrt{6}}}{{3}} )

So, the critical points are ( x = \frac{{9 + 2\sqrt{6}}}{{3}} ) and ( x = \frac{{9 - 2\sqrt{6}}}{{3}} ).

Now, we find the second derivative:

( f''(x) = 6x - 18 )

Evaluate ( f''(x) ) at the critical points:

For ( x = \frac{{9 + 2\sqrt{6}}}{{3}} ), ( f''\left(\frac{{9 + 2\sqrt{6}}}{{3}}\right) > 0 ), so it's a local minimum.

For ( x = \frac{{9 - 2\sqrt{6}}}{{3}} ), ( f''\left(\frac{{9 - 2\sqrt{6}}}{{3}}\right) < 0 ), so it's a local maximum.

Therefore, the local extrema of the function are:

Local minimum at ( x = \frac{{9 + 2\sqrt{6}}}{{3}} )

Local maximum at ( x = \frac{{9 - 2\sqrt{6}}}{{3}} )

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7