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AfilCa [17]
3 years ago
10

(a). Consider the parabolic function f(x) = ax² +bx+c, where a ±0, b and care constants. For what values of a, band c is f

Mathematics
1 answer:
ki77a [65]3 years ago
5 0

Information about concavity is contained in the second derivative of a function. Given f(x) = ax² + bx + c, we have

f'(x) = 2ax + b

and

f''(x) = 2a

Concavity changes at a function's inflection points, which can occur wherever the second derivative is zero or undefined. In this case, since a ≠ 0, the function's concavity is uniform over its entire domain.

(i) f is concave up when f'' > 0, which occurs when a > 0.

(ii) f is concave down when f'' < 0, and this is the case if a < 0.

In Mathematica, define f by entering

f[x_] := a*x^2 + b*x + c

Then solve for intervals over which the second derivative is positive or negative, respectively, using

Reduce[f''[x] > 0, x]

Reduce[f''[x] < 0, x]

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The inverse of matrix A = \left[\begin{array}{cc}2&2&1&3\\\end{array}\right] is A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|      

The given matrix is A = \left[\begin{array}{cc}2&2&1&3\\\end{array}\right]

The inverse of a matrix is given by

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Now, determinant of matrix A is

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= (3)(2) - (2)(1)

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Now, adjoint A will be

A_{11} = 3

A_{12} = -1

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AdjA = \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|

Now, substituting the values of |A| and AdjA in equation (1), we get

A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|        

Therefore, A^{-1} = \frac{1}{4} \left|\begin{array}{cc}3&-2&-1&2\\\end{array}\right|      

Know more about Inverse of Matrix: -brainly.com/question/12442362

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