Question

Given that the graph of \(f\) passes through the point \(\left( {1,6} \right)\) and that the slope of its tangent line the slope of its tangent line at \(\left( {x,f\left( x \right)} \right)\) is \(2x + 1\), find \(f\left( 2 \right)\).

Short answer

Required function is \(f\left( x \right) = {x^2} + x + 4\) and \(f\left( 2 \right) = 10\).

Step-by-step solution

Find the function from the given conditions

Here given that, the slope of tangent line at\(\left( {x,f\left( x \right)} \right)\)is\(2x + 1\).

So, we can write\(f'\left( x \right) = 2x + 1\).

Now, general anti-derivative of\(f'\left( x \right) = 2x + 1\)is,

\(\begin{aligned}{c}f\left( x \right) &= 2\frac{{{x^2}}}{2} + x + C\\ &= {x^2} + x + C{\rm{ }}.....\left( 1 \right)\end{aligned}\)

Now, given that graph of\(f\left( x \right)\)is pass through\(\left( {1,6} \right)\).

\( \Rightarrow f\left( 1 \right) = 6\)

Put above value in equation (1)

\(\begin{aligned}{l} \Rightarrow 6 &= {\left( 1 \right)^2} + 1 + C\\ \Rightarrow C &= 4\end{aligned}\)

So, required function is \(f\left( x \right) = {x^2} + x + 4\)

Find \(f\left( 2 \right)\)

Here we have\(f\left( x \right) = {x^2} + x + 4\).

To find\(f\left( 2 \right)\)we have to put\(x = 2\)in\(f\left( x \right) = {x^2} + x + 4\).

\( \Rightarrow f\left( 2 \right) = {\left( 2 \right)^2} + 2 + 4 = 10\)

Hence, \(f\left( 2 \right) = 10\).