Question

If the tangent line to \(y = f\left( x \right)\) at \(\left( {{\bf{4}},{\bf{3}}} \right)\)passes through the point \(\left( {{\bf{0}},{\bf{2}}} \right)\), find \(f\left( {\bf{4}} \right)\) and \(f'\left( {\bf{4}} \right)\).

Short answer

The value of \(f\left( 4 \right)\) is 3.

The value of \(f'\left( 4 \right)\) is \(\frac{1}{4}\).

Step-by-step solution

Step 1:Find the value of \(f\left( {\bf{4}} \right)\)

It is given that the point \(\left( {4,3} \right)\) lies on the curve \(y = f\left( x \right)\). The point \(\left( {x,y} \right)\) is similar to \(\left( {x,f\left( x \right)} \right)\). So, \(\left( {x,f\left( x \right)} \right) \equiv \left( {4,3} \right)\).

Therefore, \(f\left( 4 \right) = 3\).

So, the value of \(f\left( 4 \right)\) is 3.

Find the value of \(f'\left( {\bf{4}} \right)\)

The tangent line at \(\left( {4,3} \right)\) is passing through the point \(\left( {0,2} \right)\). So, the slope of the line joining\(\left( {4,3} \right)\) and \(\left( {0,2} \right)\) is \(f'\left( 4 \right)\).

The slope of the line \(\left( {4,3} \right)\) and \(\left( {0,2} \right)\) can be calculated as follows:

\(\begin{aligned}f'\left( 4 \right) &= \frac{{3 - 2}}{{4 - 0}}\\ &= \frac{1}{4}\end{aligned}\)

So, the value of \(f'\left( 4 \right)\) is \(\frac{1}{4}\).