Question

for the function ( f ), ( f(x)=2x + 1 ) and ( f(1)=4 ). what is the approximation for ( f(1.2) ) found by using the line tangent to the graph of ( f ) at ( x = 1 )?
a 0.6
b 3.4
c 4.2
d 4.6
e 4.64

Explanation:

Step1: Find the slope of the tangent line

The slope of the tangent line at \(x = 1\) is \(f^{\prime}(1)\). Given \(f^{\prime}(x)=2x + 1\), then \(f^{\prime}(1)=2\times1+1=3\).

Step2: Use the point - slope form of a line

The point - slope form of a line is \(y - y_0=m(x - x_0)\). Here \(x_0 = 1\), \(y_0=f(1) = 4\) and \(m = 3\). So the equation of the tangent line is \(y-4=3(x - 1)\), which simplifies to \(y=3x+1\).

Step3: Approximate \(f(1.2)\)

Substitute \(x = 1.2\) into the tangent line equation. \(y=3\times1.2 + 1=3.6+1=4.6\).

Answer:

D. 4.6