For what values of x is #f(x)=(2x-2)(x-4)(x-3)# concave or convex?

Answer 1

Concave where #x < 2.67# and Convex where #x > 2.67#

It is concave if the second derivative is less than zero and convex if the second derivative is greater than zero.
# f(x)=(2x−2)(x−4)(x−3) ; f(x)=(2x−2)(x^2−7x + 12) ; f(x) = 2x^3 – 14x^2 +24x -2x^2 +14x -24 ; f(x) = 2x^3 - 16x^2 + 38x – 24#
First Derivative: #6x^2 – 32x + 38# ; Second derivative: #12x – 32#

Concave where #12x – 32 < 0# and Convex where #12x – 32 > 0#
Concave where #x < 2.67# and Convex where #x > 2.67#

An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes. Inflection points may be stationary points, but are not local maxima or local minima. For example, for the curve y = x^3, the point x=0 is an inflection point.
First Derivative Test

Suppose f(x) is continuous at a stationary point x_0.
1. If f^'(x)>0 on an open interval extending left from x_0 and f^'(x)<0 on an open interval extending right from x_0, then f(x) has a local maximum (possibly a global maximum) at x_0.
2. If f^'(x)<0 on an open interval extending left from x_0 and f^'(x)>0 on an open interval extending right from x_0, then f(x) has a local minimum (possibly a global minimum) at x_0.
3. If f^'(x) has the same sign on an open interval extending left from x_0 and on an open interval extending right from x_0, then f(x) has an inflection point at x_0.
CITE THIS AS:
Weisstein, Eric W. "First Derivative Test." From MathWorld--A Wolfram Web Resource. https://tutor.hix.ai
REFERENCES:
Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 14, 1972.
Referenced on Wolfram|Alpha: First Derivative Test

Second Derivative Test
Suppose f(x) is a function of x that is twice differentiable at a stationary point x_0.
1. If f^('')(x_0)>0, then f has a local minimum at x_0.
2. If f^('')(x_0)<0, then f has a local maximum at x_0.
The extremum test gives slightly more general conditions under which a function with f^('')(x_0)=0 is a maximum or minimum.
If f(x,y) is a two-dimensional function that has a local extremum at a point (x_0,y_0) and has continuous partial derivatives at this point, then f_x(x_0,y_0)=0 and f_y(x_0,y_0)=0. The second partial derivatives test classifies the point as a local maximum or local minimum.
Define the second derivative test discriminant as
D = f_(xx)f_(yy)-f_(xy)f_(yx)
= f_(xx)f_(yy)-f_(xy)^2.
Then
1. If D>0 and f_(xx)(x_0,y_0)>0, the point is a local minimum.
2. If D>0 and f_(xx)(x_0,y_0)<0, the point is a local maximum.
3. If D<0, the point is a saddle point.
4. If D=0, higher order tests must be used.
REFERENCES:
Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 14, 1972.
Thomas, G. B. Jr. and Finney, R. L. "Maxima, Minima, and Saddle Points." §12.8 in Calculus and Analytic Geometry, 8th ed. Reading, MA: Addison-Wesley, pp. 881-891, 1992.
Referenced on Wolfram|Alpha: Second Derivative Test
CITE THIS AS:
Weisstein, Eric W. "Second Derivative Test." From MathWorld--A Wolfram Web Resource. https://tutor.hix.ai

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Answer 2

To determine where the function ( f(x) = (2x - 2)(x - 4)(x - 3) ) is concave or convex, you need to find the second derivative of the function and analyze its sign.

  1. Find the first derivative of ( f(x) ) using the product rule:

[ f'(x) = (2x - 2)(x - 4) + (2x - 2)(x - 3) + (x - 4)(x - 3) ]

  1. Simplify the expression:

[ f'(x) = (2x^2 - 8x - 2x + 8) + (2x^2 - 6x - 2x + 6) + (x^2 - 3x - 4x + 12) ]

[ f'(x) = 2x^2 - 10x + 8 + 2x^2 - 8x + 6 + x^2 - 7x + 12 ]

[ f'(x) = 5x^2 - 25x + 26 ]

  1. Find the second derivative of ( f(x) ):

[ f''(x) = 10x - 25 ]

  1. Set ( f''(x) ) equal to zero to find critical points:

[ 10x - 25 = 0 ] [ 10x = 25 ] [ x = 2.5 ]

  1. Test the intervals determined by the critical point using the second derivative test:
  • ( f''(x) > 0 ) implies concave up.
  • ( f''(x) < 0 ) implies concave down.

Test ( x = 2 ): [ f''(2) = (10 \times 2) - 25 = 5 > 0 ] Therefore, the function is concave up when ( x < 2.5 ).

Test ( x = 3 ): [ f''(3) = (10 \times 3) - 25 = 5 > 0 ] Therefore, the function is concave up when ( 2.5 < x < \infty ).

Test ( x = 4 ): [ f''(4) = (10 \times 4) - 25 = 15 > 0 ] Therefore, the function is concave up when ( 2.5 < x < \infty ).

Thus, the function is concave up for ( x \in (2.5, \infty) ).

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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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