The values of constants a and b for which the function, y=a logc x+b x^2+x has a local minimum a

The values of constants a and b for which the function, y=a...

Key Concepts

Critical Points and Derivative Conditions

In calculus, local minima and maxima occur at critical points where the first derivative of the function is zero. Setting the derivative equal to zero at specified points provides equations for any unknown constants. This principle is used to tailor the function to have desired extremum behaviors at particular values of the variable.

Second Derivative Test for Extrema

After finding critical points using the first derivative, the second derivative test helps to classify their nature. If the second derivative at a critical point is positive, the point is a local minimum; if it is negative, the point is a local maximum. This test is essential in determining the sign of the curvature at the critical points, which in turn ensures the correctness of the solution.

Differentiation of Logarithmic Functions

The derivative of a logarithmic function with a general base is given by d/dx (log_c x) = 1/(x ln(c)), where ln(c) is the natural logarithm of the base. This formula is crucial when differentiating functions that involve logarithms, and its proper application is necessary to set up accurate equations for solving problems involving mixed logarithmic and polynomial terms.

Solving Systems of Equations

When a function is subject to multiple conditions, such as having a local minimum at one point and a local maximum at another, it leads to a system of equations arising from the derivative conditions. Solving these simultaneous equations allows for the determination of unknown parameters in the function, ensuring that all specified conditions are satisfied.

Transcript

00:01 Okay, so we want to find a, b, and c, such that this function has a local minimum x equals 3 and the local max at negative 1, negative 2.

00:42 Yeah, so where are we going to start here? well, we know, first of all, that if we have a local minimum and a local max here, then then let's take the derivative and that'll tell us something about the derivative so bx plus c times 2x minus x squared plus a times b we said that equal to zero then we just want this numerator to be so let's see bx c let's go and multiply that by 2x so 2 bx squared plus 2cx minus b x squared minus a times b should be 0 okay but we know that our two roots are negative 1 and 3 okay so we have let's simplify this b x squared plus 2c x minus ab is 0 but we know the solutions are negative 1 and 3.

02:21 So if we plug in negative 1, we get negative b, positive b, negative 1 squared minus 2c minus a b is 0.

02:34 And if we plug in 3, we get 9 times b plus 3 times 2, 3 times 2 times c, so 6c minus ab is 0.

02:51 Okay.

02:53 Natural thing to do is just to subtract these equations so i'm going to subtract this one from this one so that i'll give me 8b and plus 8c is 0 okay so that tells me that b equals negative c okay so far so good and so if i plug in b equals minus c i'll have minus c minus c minus 2c plus a c is zero or a c equals 3c so then either a is 3 or c is 0 good so if c is zero that's going to be a big time problem and why is that a problem because if c is zero then b is zero and if b and c are both zero then the the entire numerator is zero, so the entire function is undefined.

04:13 So that's going to be a big problem.

04:14 So we do not have that c is equal to zero.

04:17 We have the a is equal to 3.

04:25 So how are we going to find being c? well, let's look...

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