Math Problem Statement

7,8,9

The problems from the image involve demand functions, cost functions, square root function differentiation, and polynomial function differentiation.

Solution

Let's break down and solve the problems from the image.

Question 7:

(a) The demand function D(P)=abPD(P) = a - bP, find dD(P)dP\frac{dD(P)}{dP}.

This is a basic differentiation problem. Differentiating D(P)=abPD(P) = a - bP with respect to PP:

dD(P)dP=ddP(abP)=0b=b\frac{dD(P)}{dP} = \frac{d}{dP}(a - bP) = 0 - b = -b

(b) The cost function C(x)=p+qx2C(x) = p + qx^2, find C(x)C'(x).

Differentiating C(x)=p+qx2C(x) = p + qx^2 with respect to xx:

C(x)=ddx(p+qx2)=0+2qx=2qxC'(x) = \frac{d}{dx}(p + qx^2) = 0 + 2qx = 2qx

Question 8:

(a) Show that (x+hx)(x+h+x)=h(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x}) = h.

This expression can be simplified using the difference of squares:

(x+hx)(x+h+x)=(x+h)2(x)2=(x+h)x=h(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x}) = (\sqrt{x + h})^2 - (\sqrt{x})^2 = (x + h) - x = h

(b) If f(x)=xf(x) = \sqrt{x}, show that f(x+h)f(x)h=1x+h+x\frac{f(x + h) - f(x)}{h} = \frac{1}{\sqrt{x + h} + \sqrt{x}}.

We need to compute x+hxh\frac{\sqrt{x + h} - \sqrt{x}}{h}. Multiply the numerator and denominator by x+h+x\sqrt{x + h} + \sqrt{x}:

x+hxhx+h+xx+h+x=(x+h)2(x)2h(x+h+x)=(x+h)xh(x+h+x)\frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(\sqrt{x + h})^2 - (\sqrt{x})^2}{h(\sqrt{x + h} + \sqrt{x})} = \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} =hh(x+h+x)=1x+h+x= \frac{h}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}}

(c) Use the result in part (b) to show that for x>0x > 0, f(x)=xf(x) = \sqrt{x} implies f(x)=12xf'(x) = \frac{1}{2\sqrt{x}}.

The derivative of f(x)=xf(x) = \sqrt{x} is the limit as h0h \to 0 of the expression in part (b):

f(x)=limh0x+hxh=12xf'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} = \frac{1}{2\sqrt{x}}

(d) Show that the result could also be written as ddxx1/2=12x1/2\frac{d}{dx} x^{1/2} = \frac{1}{2}x^{-1/2}.

We know that f(x)=x1/2f(x) = x^{1/2}. Using power rule for differentiation:

ddxx1/2=12x1/21=12x1/2\frac{d}{dx} x^{1/2} = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2}

Question 9:

(a) If f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d, show that

f(x+h)f(x)h=3ax2+2bx+c+3axh+ah2+bh\frac{f(x + h) - f(x)}{h} = 3ax^2 + 2bx + c + 3axh + ah^2 + bh

First, expand f(x+h)=a(x+h)3+b(x+h)2+c(x+h)+df(x + h) = a(x + h)^3 + b(x + h)^2 + c(x + h) + d:

f(x+h)=a(x3+3x2h+3xh2+h3)+b(x2+2xh+h2)+c(x+h)+df(x + h) = a(x^3 + 3x^2h + 3xh^2 + h^3) + b(x^2 + 2xh + h^2) + c(x + h) + d =ax3+3ax2h+3axh2+ah3+bx2+2bxh+bh2+cx+ch+d= ax^3 + 3ax^2h + 3axh^2 + ah^3 + bx^2 + 2bxh + bh^2 + cx + ch + d

Now subtract f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d:

f(x+h)f(x)=3ax2h+3axh2+ah3+2bxh+bh2+chf(x + h) - f(x) = 3ax^2h + 3axh^2 + ah^3 + 2bxh + bh^2 + ch

Now divide by hh:

f(x+h)f(x)h=3ax2+3axh+ah2+2bx+bh+c\frac{f(x + h) - f(x)}{h} = 3ax^2 + 3axh + ah^2 + 2bx + bh + c

(b) Hence, f(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c.

As h0h \to 0, the terms involving hh vanish, and we are left with:

f(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c


Would you like further details on any part of these solutions?

Related Questions:

  1. How do you find the second derivative of f(x)=xf(x) = \sqrt{x}?
  2. What is the interpretation of a demand function in economics?
  3. How do you apply the product rule to differentiate a product of two functions?
  4. Can you find the critical points of f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d?
  5. What is the difference between marginal cost and marginal revenue in economic terms?

Tip:

Always check the domain of the function when differentiating, especially with square roots and logarithms, to avoid undefined regions.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Demand and cost functions
Algebraic manipulation (difference of squares)
Power rule

Formulas

Demand function: D(P) = a - bP
Cost function: C(x) = p + qx^2
Derivative of a square root function: d(√x)/dx = 1 / (2√x)
Derivative of a power: d(x^n)/dx = nx^(n-1)
Difference of squares: (A - B)(A + B) = A^2 - B^2

Theorems

Difference of squares
Power rule for differentiation

Suitable Grade Level

Grades 10-12