Question

consider the following.

$f(x)=3x^{2}-7$

find the following values of the function.

$f(2 + delta x)=$

$f(2)=$

find the slope of the tangent line to the graph of the function at the point $(2,5)$.

Explanation:

Step1: Substitute \( x=2+\Delta x \) into \( f(x) \)

\( f(2+\Delta x) = 3(2+\Delta x)^2 - 7 \)

Step2: Expand \( (2+\Delta x)^2 \)

\( (2+\Delta x)^2 = 4 + 4\Delta x + (\Delta x)^2 \)

Step3: Simplify the expression

\( 3(4 + 4\Delta x + (\Delta x)^2) - 7 = 3(\Delta x)^2 + 12\Delta x + 5 \)

Step4: Substitute \( x=2 \) into \( f(x) \)

\( f(2) = 3(2)^2 - 7 \)

Step5: Calculate \( f(2) \)

\( 3(4) - 7 = 5 \)

Step6: Find derivative \( f'(x) \)

\( f'(x) = 6x \)

Step7: Evaluate \( f'(2) \)

\( f'(2) = 6(2) = 12 \)

Answer:

\( f(2 + \Delta x) = 3(\Delta x)^2 + 12\Delta x + 5 \)
\( f(2) = 5 \)
The slope of the tangent line is 12