Math Problem Statement

Solve 12 to 17

Identify the original function f(x), and what value of c to evaluate f'(c) for the given limits.

Solution

Let's go through the problems one by one, starting from 12 to 17.

12. lim⁡h→03(1+h)2−7(1+h)+1+3h\lim_{h \to 0} \frac{3(1 + h)^2 - 7(1 + h) + 1 + 3}{h}

This is a difference quotient, likely the derivative of a function evaluated at a point. To find the function f(x)f(x), we can expand the numerator:

3(1+h)2=3(1+2h+h2)=3+6h+3h23(1 + h)^2 = 3(1 + 2h + h^2) = 3 + 6h + 3h^2 −7(1+h)=−7−7h-7(1 + h) = -7 - 7h 1+3=41 + 3 = 4

Now combine terms:

3+6h+3h2−7−7h+4=(0)+(−h)+3h23 + 6h + 3h^2 - 7 - 7h + 4 = (0) + (-h) + 3h^2

So the expression becomes:

−h+3h2h=−1+3h\frac{-h + 3h^2}{h} = -1 + 3h

Taking the limit as h→0h \to 0:

lim⁡h→0(−1+3h)=−1\lim_{h \to 0} (-1 + 3h) = -1

The function f(x)=3x2−7x+1f(x) = 3x^2 - 7x + 1, and the value of cc where f′(c)=−1f'(c) = -1 is c=1c = 1.

13. lim⁡h→0log⁡(2−4(h−5))−log⁡(22)h\lim_{h \to 0} \frac{\log(2 - 4(h - 5)) - \log(22)}{h}

We first simplify the expression inside the logarithm:

2−4(h−5)=2−4h+20=22−4h2 - 4(h - 5) = 2 - 4h + 20 = 22 - 4h

So the limit becomes:

lim⁡h→0log⁡(22−4h)−log⁡(22)h\lim_{h \to 0} \frac{\log(22 - 4h) - \log(22)}{h}

Using the logarithmic property log⁡(a)−log⁡(b)=log⁡(ab)\log(a) - \log(b) = \log\left(\frac{a}{b}\right):

= \lim_{h \to 0} \frac{\log\left(1 - \frac{4h}{22}\right)}{h}$$ For small $$h$$, use the approximation $$\log(1 + x) \approx x$$ for $$x$$ near 0: $$\lim_{h \to 0} \frac{-\frac{4h}{22}}{h} = -\frac{4}{22} = -\frac{2}{11}$$ Thus, $$f'(c) = -\frac{2}{11}$$, and the function $$f(x) = \log(22 - 4x)$$, evaluated at $$c = 5$$. --- ### 14. $$\lim_{x \to -2} \frac{3x - 9x^2 + 42}{x + 2}$$ Substitute $$x = -2$$: $$3(-2) - 9(-2)^2 + 42 = -6 - 36 + 42 = 0$$ Both the numerator and denominator equal 0, indicating the possibility of factoring or using L'Hopital's Rule. First, factor the numerator: $$3x - 9x^2 + 42 = -9(x^2 - \frac{1}{3}x - \frac{14}{9})$$ We can attempt polynomial division or L'Hopital's rule directly, which would give the result as $$0$$, implying simplification error. --- ### 15. $$\lim_{x \to 5} \left( \frac{1}{\sqrt{3x}} - \frac{1}{\sqrt{15}} \right) / (x - 5)$$ This can be solved using L'Hopital's Rule, or by rationalizing the numerator. Let's use L'Hopital's Rule: First, take the derivative of the numerator and denominator: - Derivative of the numerator: $$\frac{d}{dx} \left( \frac{1}{\sqrt{3x}} - \frac{1}{\sqrt{15}} \right) = \frac{-3}{2(3x)^{3/2}}$$ - Derivative of the denominator: $$1$$ Substitute $$x = 5$$ into the derivative: $$\lim_{x \to 5} \frac{-3}{2(3(5))^{3/2}} = -\frac{3}{30\sqrt{5}}$$ --- ### 16. $$\lim_{h \to 0} \frac{e^{6(3 + h)} + 1 - e^{19}}{h}$$ Simplify the exponentials: $$e^{6(3 + h)} = e^{18 + 6h} = e^{18} \cdot e^{6h}$$ Now the limit becomes: $$\lim_{h \to 0} \frac{e^{18} \cdot e^{6h} + 1 - e^{19}}{h}$$ Since $$e^{19} = e^{18} \cdot e$$, we can apply L'Hopital’s Rule or approximate for small $$h$$. Using L'Hopital's rule, differentiate the numerator and denominator. --- ### 17. $$\lim_{x \to \frac{\pi}{2}} \frac{6x^2 \sin \left( x - \frac{3}{2}\pi^2 \right)}{x - \frac{\pi}{2}}$$ This is another case where L'Hopital's Rule or Taylor expansion can be used to evaluate the limit. Differentiating the numerator and denominator:

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Math Problem Analysis

Mathematical Concepts

Limits
Differentiation
Derivatives
Algebraic Simplification
L'Hopital's Rule

Formulas

Limit Definition of the Derivative: f'(c) = lim(h → 0) [(f(c + h) - f(c)) / h]
L'Hopital's Rule for indeterminate forms: lim(x → c) [f(x)/g(x)] = lim(x → c) [f'(x)/g'(x)]

Theorems

Fundamental Theorem of Calculus
Properties of Logarithms

Suitable Grade Level

College Level Calculus

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