In differentiating the quadratic function f(x) = 2x^2 + 5x from first principles, which of… — JAMB UTME

In differentiating the quadratic function $f(x) = 2x^2 + 5x$ from first principles, which of the following expressions represents the fully simplified difference quotient $\frac{f(x+h) - f(x)}{h}$ prior to evaluating the limit as $h \to 0$ ?

$4x + 2h + 5$ Answer
B
$4x + h + 5$
C
$4x - 5 + 2h$
D
$2x + 2h + 5$

Answer

The fully simplified difference quotient is

4x + 2h + 5

Expanding

f(x+h) = 2(x+h)^2 + 5(x+h)

gives

2x^2 + 4xh + 2h^2 + 5x + 5h

. Subtracting

f(x) = 2x^2 + 5x

leaves

4xh + 2h^2 + 5h

. Dividing this expression term-by-term by

h

yields

4x + 2h + 5

Step-by-Step Solution

Evaluate

f(x+h)

for the function

f(x) = 2x^2 + 5x

f(x+h) = 2(x+h)^2 + 5(x+h) = 2(x^2 + 2xh + h^2) + 5x + 5h = 2x^2 + 4xh + 2h^2 + 5x + 5h

To apply first principles, we must replace every instance of

x

with

(x+h)

and expand completely.

Form the numerator of the difference quotient,

f(x+h) - f(x)

f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 5x + 5h) - (2x^2 + 5x) = 4xh + 2h^2 + 5h

Subtracting

f(x)

cancels out the terms

2x^2

and

5x

that do not contain

h

Divide the result by

h

to get the difference quotient.

\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5

Factoring out

h

from the numerator and canceling it with the denominator simplifies the quotient.

Key Concept

Differentiation from First Principles (Difference Quotient)

Estimated Time:1m 30s

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