How do you find the critical points to sketch the graph #g(x)=x^4-8x^2-10#?

Answer 1

Start by finding the first derivative.

#g'(x) = 4x^3 - 16x#
Now find the critical numbers, which will occur when the derivative #=# #0#.
#0 = 4x^3 - 16x#
#0 = 4x(x^2 - 4)#
#0 = 4x(x + 2)(x - 2)#
#x = 0, 2, and -2#

We now determine the second derivative to find the points of inflection.

#g''(x) = 12x^2 - 16#
Once again, we set the derivative to #0# and solve (except this time it'll be the second derivative).
#0 = 12x^2 - 16#
#0 = 4(3x^2 - 4)#
#x = +- 2/sqrt(3)#
Now let's revert our attention back to the 1st derivative. We must determine the intervals of increase decrease. Quite simply, if #g'(a) > 0#, then #g(x)# is increasing at that point, and if #g'(a) < 0#, then #g(x)# is decreasing at that point.

Select test points between the critical points.

Test point 1: #x = -1#
#g'(-1) = 4(-1)^3 - 16(-1) = -4 + 16 = 12#
Since this is positive, the function is increasing on #(-2, 0)#.
Test point 2: #x = 1#
#g'(1) = 4(1)^3 - 16(1) = 4 - 16 = -12#
Since this is negative, the function is decreasing on #(0, 2)#
Test point 3: #x = 3#
#g'(3) = 4(3)^3 - 16(3) = 60#
Since this is positive, the function is increasing on #(2, oo)#.
Test point 4: #x = -3#
#g'(-3) = 4(-3)^3 - 16(-3) = -60#
Since this is negative, the function is decreasing on #(-oo, -2)#.
Now we go back to the second derivative to check the concave up/concave down intervals. If #g''(a) > 0#, then #g(x)# is concave up at that point, and if #g''(a) < 0#, then #g(x)# is concave down at that point.

We will once again select test points.

Test point 1: #x = -3#
#g''(-3) = 12(-3)^2 - 16 = 92#
Since this is positive, the function is concave up on #(-oo, -2/sqrt(3))#.
Test point 2: #x = 0#
#g''(0) = 12(0)^2 - 16 = -16#
Since this is negative, the function is concave down on #(-2/sqrt(3), 2/sqrt(3))#.
Test point 3: #x = 3#
#g''(3) = 12(3)^3 - 16 = 92#
Since this is positive, the function is concave up on #(2/sqrt(3), oo)#.

It's true that you could just have found the next intervals of concavity and increasing/decreasing after the first by following the pattern of positive-negative-positive/negative-positive-negative, depending on the first interval, but I wanted to show you this method to make it as clear as possible.

The last thing I would like to discuss before graphing is intercepts. First, for the x-intercepts.

#0 = x^4 - 8x^2 - 10#
We let #u = x^2#.
#0 = u^2 - 8u - 10#
#u = (-(-8) +- sqrt((-8)^2 - 4 xx 1 xx -10))/(2 xx 1)#
#u = (8 +- 2sqrt(26))/2#
#u = 4 +- sqrt(26)#
We now revert to #x#.
#x^2 = 4 +- sqrt(26)#
#x = +- sqrt(4 +- sqrt(26))#
But since the value under the #√# must always be positive, the #+-# sign should become a negative a #+#.
#x = +- sqrt(4 + sqrt(26))#

Finally, as for the y-intercept, we have:

#g(0) = 0^4 - 8(0)^2 - 10= -10#

We now trace the following graph putting all of the previous elements together.

graph{x^4 - 8x^2 - 10 [-58.5, 58.5, -29.27, 29.3]}

Hopefully this helps!

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Answer 2

To find the critical points of ( g(x) = x^4 - 8x^2 - 10 ), follow these steps:

  1. Compute the first derivative ( g'(x) ).

  2. Set ( g'(x) = 0 ) to find the values of ( x ).

  3. Solve for ( x ) to determine the critical points.

  4. Compute ( g'(x) ): [ g'(x) = 4x^3 - 16x ]

  5. Set ( g'(x) = 0 ): [ 4x^3 - 16x = 0 ]

  6. Factor out the common term: [ 4x(x^2 - 4) = 0 ]

Solve for ( x ): [ 4x = 0 \Rightarrow x = 0 ] [ x^2 - 4 = 0 \Rightarrow x = 2, x = -2 ]

The critical points are ( x = 0, x = 2, ) and ( x = -2 ).

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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.

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