If f(x)=(x^(2)+1)^(3), what is lim_(x rarr-1)(f(x)-f(-1))/(x+1)?

Substitute x = -1: First, substitute x = -1 into f(x) to find f(-1). f(-1) = ((-1)^2 + 1)^3 = (1 + 1)^3 = 2^3 = 8.Simplify expression: Now, simplify the expression (f(x) - f(-1)) / (x + 1). (f(x) - f(-1)) / (x + 1) = ((x^2 + 1)^3 - 8) / (x + 1).Factorize the numerator: To find the limit as x approaches -1, we need to simplify the numerator. Notice that x^2 + 1 = 2 when x = -1, so we can factorize the numerator using the difference of cubes: (x^2 + 1)^3 - 8 = (2)^3 - 8 = 8 - 8 = 0. This shows that the numerator is zero at x = -1.Check denominator: Since the numerator is zero at x = -1, we need to check if the denominator also becomes zero to avoid division by zero. At x = -1, the denominator x + 1 = -1 + 1 = 0.Apply L'Hopital's Rule: Both the numerator and denominator are zero at x = -1, indicating a 0/0 form. We apply L'Hopital's Rule: Differentiate the numerator and the denominator separately. Numerator: d/dx[(x^2 + 1)^3] = 3(x^2 + 1)^2 * 2x = 6x(x^2 + 1)^2. Denominator: d/dx[x + 1] = 1.Find limit of derivatives: Now, find the limit of the derivatives as x approaches -1: lim_(x -> -1) [6x(x^2 + 1)^2 / 1] = 6(-1)(1 + 1)^2 = 6(-1)(2)^2 = -24.

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If
f(x)=(x^(2)+1)^(3), what is
lim_(x rarr-1)(f(x)-f(-1))/(x+1)?

If $f(x)=\left(x^{2}+1\right)^{3}$ , what is $\lim _{x \rightarrow-1} \frac{f(x)-f(-1)}{x+1} ?$

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Calculus

Evaluate definite integrals using the chain rule

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Q. If

f(x)=\left(x^{2}+1\right)^{3}

, what is

\lim _{x \rightarrow-1} \frac{f(x)-f(-1)}{x+1} ?

Substitute $x = -1$ : First, substitute $x = -1$ into $f(x)$ to find $f(-1)$ . $\newline$ $f(-1) = ((-1)^2 + 1)^3 = (1 + 1)^3 = 2^3 = 8$ .
Simplify expression: Now, simplify the expression $(f(x) - f(-1)) / (x + 1)$ . $(f(x) - f(-1)) / (x + 1) = ((x^2 + 1)^3 - 8) / (x + 1)$ .
Factorize the numerator: To find the limit as $x$ approaches $-1$ , we need to simplify the numerator. Notice that $x^2 + 1 = 2$ when $x = -1$ , so we can factorize the numerator using the difference of cubes: $(x^2 + 1)^3 - 8 = (2)^3 - 8 = 8 - 8 = 0$ . This shows that the numerator is zero at $x = -1$ .
Check denominator: Since the numerator is zero at $x = -1$ , we need to check if the denominator also becomes zero to avoid division by zero. $\newline$ At $x = -1$ , the denominator $x + 1 = -1 + 1 = 0$ .
Apply L'Hopital's Rule: Both the numerator and denominator are zero at $x = -1$ , indicating a $0/0$ form. We apply L'Hopital's Rule: $\newline$ Differentiate the numerator and the denominator separately. $\newline$ Numerator: $\frac{d}{dx}[(x^2 + 1)^3] = 3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2$ . $\newline$ Denominator: $\frac{d}{dx}[x + 1] = 1$ .
Find limit of derivatives: Now, find the limit of the derivatives as $x$ approaches $-1$ : $\lim_{x \to -1} \left[\frac{6x(x^2 + 1)^2}{1}\right] = 6(-1)(1 + 1)^2 = 6(-1)(2)^2 = -24$ .