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

The curve is concave downwards at

The curve is concave upwards at

The curve is concave upwards at

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

#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

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

At

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

Since the second derivative is less than zero, the curve is concave downwards at

The value of the function is -

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

The curve is concave downwards at

At

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

Since the second derivative is greater than zero, the curve is concave upwards at

The value of the function is -

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

The curve is concave upwards at

At

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

Since the second derivative is greater than zero, the curve is concave upwards at

The value of the function is -

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

The curve is concave upwards at

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