Is #f(x)=x^3-x^2+x-4# concave or convex at #x=-1#?

Answer 1

#"concave at " x=-1#

To determine if a function is concave/convex at f ( a), we require to find the value of f'' ( a)

#• " If " f''(a)>0" then "f(x)" is convex at x=a"#
#• " If " f''(a)<0" then " f(x)" is concave at x=a"#
#f(x)=x^3-x^2+x-4#
#rArrf'(x)=3x^2-2x+1#
#rArrf''(x)=6x-2#
#"and "f''(-1)=-6-2=-8<0#
#rArrf(x)" is concave at "x=-1# graph{x^3-x^2+x-4 [-10, 10, -5, 5]}
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Answer 2

To determine whether the function ( f(x) = x^3 - x^2 + x - 4 ) is concave or convex at ( x = -1 ), we need to examine the second derivative of the function.

First, find the first derivative of ( f(x) ) with respect to ( x ), which gives ( f'(x) = 3x^2 - 2x + 1 ).

Now, find the second derivative of ( f(x) ), which is ( f''(x) = 6x - 2 ).

Substitute ( x = -1 ) into the second derivative: ( f''(-1) = 6(-1) - 2 = -8 ).

Since ( f''(-1) < 0 ), the function is concave at ( x = -1 ).

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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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