Find Where the Mean Value Theorem is Satisfied f(x)=4x^2-8x+3 , [-1,3]

Solution:

Rephrased Problem-Solving Process:

Step 1: The Mean Value Theorem (MVT) states that for a function $f$ that is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists at least one number $c$ in $(a,b)$ such that $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$, which equates the instantaneous rate of change at $c$ with the average rate of change over $[a,b]$.

Step 2: Examine the continuity of $f(x)=4x^2-8x+3$.

Step 2.1: The function's domain is all real numbers, as there are no values for which it is undefined. In interval notation, this is $(-\mathrm{\infty },\mathrm{\infty })$, and in set-builder notation, $\{x|x\in \mathbb{R}\}$.

Step 2.2: $f(x)$ is continuous over the interval $[-1,3]$.

Step 3: Determine the derivative of $f(x)$.

Step 3.1: Calculate the first derivative of $f(x)$.

Step 3.1.1: Apply the Sum Rule to find the derivative of each term: $\frac{d}{dx}[4x^2]+\frac{d}{dx}[-8x]+\frac{d}{dx}[3]$.

Step 3.1.2: Compute $\frac{d}{dx}[4x^2]$.

Step 3.1.2.1: Using the constant multiple rule, the derivative is $4\frac{d}{dx}[x^2]$.

Step 3.1.2.2: Apply the Power Rule, which gives $4(2x)$.

Step 3.1.2.3: Multiply to get $8x$.

Step 3.1.3: Compute $\frac{d}{dx}[-8x]$.

Step 3.1.3.1: The derivative is $-8\frac{d}{dx}[x]$.

Step 3.1.3.2: Apply the Power Rule to get $-8(1)$.

Step 3.1.3.3: The result is $-8$.

Step 3.1.4: Use the Constant Rule to differentiate $3$.

Step 3.1.4.1: The derivative of a constant is $0$.

Step 3.1.4.2: Combine the results to get $f^{\prime }(x)=8x-8$.

Step 4: Verify that the derivative is continuous on $(a,b)$, which is $(-1,3)$.

Step 4.1: The domain of $f^{\prime }(x)$ is all real numbers.

Step 4.2: $f^{\prime }(x)$ is continuous on $(-1,3)$.

Step 5: Since the derivative is continuous, $f(x)$ is differentiable on $(-1,3)$.

Step 6: $f(x)$ meets the MVT conditions; it is continuous on $[-1,3]$ and differentiable on $(-1,3)$.

Step 7: Evaluate $f(a)$ where $a=-1$.

Step 7.1: Substitute $x$ with $-1$: $f(-1)=4(-1{)}^2-8(-1)+3$.

Step 7.2: Simplify the expression.

Step 7.2.1: Simplify each term.

Step 7.2.1.1: Calculate $(-1{)}^2$ to get $1$.

Step 7.2.1.2: Multiply $4$ by $1$.

Step 7.2.1.3: Multiply $-8$ by $-1$ to get $8$.

Step 7.2.2: Add the numbers together.

Step 7.2.2.1: Combine $4$ and $8$.

Step 7.2.2.2: Add $12$ and $3$ to get $15$.

Step 7.2.3: The result is $f(-1)=15$.

Step 8: Solve for $x$ in $8x-8=\frac{f(b)-f(a)}{b-a}$ using $f(-1)$ and $f(3)$.

Step 8.1: Simplify the fraction on the right side of the equation.

Step 8.1.1: Simplify the numerator by subtracting $f(b)$ from $f(a)$.

Step 8.1.2: Simplify the denominator by adding $b$ to $-a$.

Step 8.1.3: Divide the numerator by the denominator to get $0$.

Step 8.2: Add $8$ to both sides of the equation to isolate $8x$.

Step 8.3: Divide both sides by $8$ to solve for $x$.

Step 8.3.1: Divide $8x$ and $8$ by $8$.

Step 8.3.2: Simplify the equation to find $x$.

Step 8.3.3: The result is $x=1$.

Step 9: A tangent line at $x=1$ is parallel to the secant line through the endpoints $a=-1$ and $b=3$.

Step 10: The Mean Value Theorem is satisfied at $x=1$ for the function $f(x)=4x^2-8x+3$ on the interval $[-1,3]$.

Knowledge Notes:

The Mean Value Theorem (MVT) is a fundamental result in calculus that relates the average rate of change of a function over an interval to its instantaneous rate of change at some point within that interval. To apply the MVT, the function must be continuous on the closed interval and differentiable on the open interval.

The Sum Rule, Power Rule, and Constant Rule are basic differentiation rules used to find the derivatives of functions. The Sum Rule allows us to differentiate a sum of functions term by term, the Power Rule applies to functions of the form $x^n$, and the Constant Rule states that the derivative of a constant is zero.

Continuity and differentiability are essential concepts in calculus. A function is continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. A function is differentiable at a point if it has a defined derivative at that point. If a function is differentiable at a point, it is also continuous at that point, but the converse is not necessarily true.