How do you use the first and second derivatives to sketch #f(x)= x^4 - 2x^2 +3#?

Answer 1

The curve is concave downwards at #(0, 3)#
The curve is concave upwards at #(1, 2)#
The curve is concave upwards at #(-1, 2)#

Given -

#y=x^4-2x^2+3#

#dy/dx=4x^3-4x#

#(d^2y)/(dx^2)=12x^2-4#

To sketch the graph, we have to find for what values of #x# the slope becomes zero. It means at those points the curve turns.

#dy/dx=0 =>4x^3-4x=0#

#4x^3-4x=0#
#4x(x^2-1)=0#
#4x=0#
#x=0#
#x^2-1=0#
#x=+-sqrt1#

#x=1#
#x=-1#

The curve turns when #x=0; x=1; x=-1#

At these points we have to decide whether the curve is concave upwards or concave downwards. For this we need the second derivatives -

At #x=0#

#(d^2y)/(dx^2)=12(0^2)-4=-4<0#

Since the second derivative is less than zero, the curve is concave downwards at #x=0#

The value of the function is -

#y=0^4-2*0^2+3=3#

The curve is concave downwards at #(0, 3)#

At #x=1#

#(d^2y)/(dx^2)=12(1^2)-4=8>0#

Since the second derivative is greater than zero, the curve is concave upwards at #x=0#

The value of the function is -

#y=1^4-2*1^2+3=2#

The curve is concave upwards at #(1, 2)#

At #x=-1#

#(d^2y)/(dx^2)=12(-1^2)-4=8>0#

Since the second derivative is greater than zero, the curve is concave upwards at #x=-1#

The value of the function is -

#y=(-1)^4-2*(-1)^2+3=2#

The curve is concave upwards at #(-1, 2)#

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Answer 2
To sketch the function \( f(x) = x^4 - 2x^2 + 3 \) using the first and second derivatives, follow these steps: 1. Find the first derivative \( f'(x) \) by differentiating \( f(x) \) with respect to \( x \). 2. Find the critical points by setting \( f'(x) = 0 \). 3. Use the first derivative test to determine the intervals where the function is increasing or decreasing. 4. Find the second derivative \( f''(x) \). 5. Use the second derivative test to determine the intervals where the function is concave up or concave down. 6. Find any points of inflection by setting \( f''(x) = 0 \). 7. Plot the critical points, points of inflection, and any other relevant points on the graph. 8. Sketch the graph by incorporating the information obtained from the first and second derivatives. Here are the detailed steps: 1. \( f'(x) = 4x^3 - 4x \) 2. Set \( f'(x) = 0 \) to find critical points: \( 4x^3 - 4x = 0 \) Factor out \( 4x \): \( 4x(x^2 - 1) = 0 \) Critical points occur when \( x = 0 \) and when \( x^2 - 1 = 0 \), which gives \( x = \pm 1 \). 3. Use the first derivative test: - When \( x < -1 \), \( f'(x) < 0 \), so \( f(x) \) is decreasing. - When \( -1 < x < 0 \), \( f'(x) > 0 \), so \( f(x) \) is increasing. - When \( 0 < x < 1 \), \( f'(x) > 0 \), so \( f(x) \) is increasing. - When \( x > 1 \), \( f'(x) > 0 \), so \( f(x) \) is increasing. 4. \( f''(x) = 12x^2 - 4 \) 5. Use the second derivative test: - \( f''(x) > 0 \) when \( x > 0 \), so \( f(x) \) is concave up. - \( f''(x) < 0 \) when \( x < 0 \), so \( f(x) \) is concave down. 6. There are no points of inflection since \( f''(x) = 0 \) has no real solutions. 7. Plot the critical points at \( x = -1, 0, 1 \). 8. Sketch the graph, noting that the function decreases from \( -\infty \) to \( -1 \), then increases from \( -1 \) to \( 0 \), decreases from \( 0 \) to \( 1 \), and increases from \( 1 \) to \( +\infty \). The function is concave down for \( x < 0 \) and concave up for \( x > 0 \).
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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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