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Chapter 3: Problem 41

A derivative formula a. Use the definition of the derivative to determine \(\frac{d}{d x}\left(a x^{2}+b x+c\right),\) where \(a, b,\) and \(c\) are constants. b. Let \(f(x)=4 x^{2}-3 x+10\) and use part (a) to find \(f^{\prime}(x)\) c. Use part (b) to find \(f^{\prime}(1)\)

Short Answer

Expert verified
Question: Find the derivative of the quadratic function \(f(x) = 4x^2 - 3x + 10\) using the definition of the derivative, and evaluate the derivative at the point \(x = 1\). Answer: The derivative of the quadratic function \(f(x)\) is \(f'(x) = 8x - 3\), and the value of the derivative at the point \(x = 1\) is \(f'(1) = 5\).

Step by step solution

01

Definition of the derivative

Recall the definition of the derivative of a function \(g(x)\): $$ g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}. $$
02

Apply the definition to the given function

We have a function \(g(x) = ax^2+bx+c\). Applying the definition of the derivative, we have $$ g'(x) = \lim_{h \to 0} \frac{(a(x+h)^2+b(x+h)+c) - (ax^2+bx+c)}{h}. $$
03

Simplify the expression inside the limit

Observe that \(c\) and \(-c\) cancels and we get $$ g'(x) = \lim_{h \to 0} \frac{a(x^2+2xh+h^2)+b(x+h) - ax^2 - bx}{h}. $$
04

Factor out \(h\) to cancel out

Factor \(h\) in the numerator: $$ g'(x) = \lim_{h \to 0} \frac{h(2ax + a h + b)}{h}. $$ Cancel out \(h\) in the numerator and denominator: $$ g'(x) = \lim_{h \to 0} (2ax + a h + b). $$
05

Evaluate the limit

As \(h\) approaches \(0\), the limit is $$ g'(x) = 2ax + b. $$
06

Apply the result to the specific function

For \(f(x)=4x^2-3x+10\), we have \(a = 4\), \(b = -3\) and \(c = 10\). So the derivative of \(f(x)\) is: $$ f'(x) = 2ax + b = 2\cdot 4x - 3 = 8x - 3. $$
07

Evaluate the derivative at the specific point

To find \(f'(1)\), we plug in \(x=1\) into our derivative formula: $$ f'(1) = 8(1) - 3 = 8 - 3 = 5. $$ So the value of the derivative \(f'(1)\) is \(5\).

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