Chapter 4.8, Problem 37E

To determine

To find: The value of $f\left(1\right)$.

Answer to Problem 37E

The value of $f\left(1\right)$ is $\underset{\_}{8}$.

Explanation of Solution

Given Data:

The function $f\left(x\right)$ passes through the point $\left(2,5\right)$ and the slope of its tangent line at $\left(x,f\left(x\right)\right)$ is $\left(3-4x\right)$.

Formula used:

The antiderivative function for the function $x^n$ is $\frac{x^{n+1}}{n+1}+C$.

Here, C is the constant.

The tangent of the function $f\left(x\right)$ is $f^{\prime }\left(x\right)$.

As the tangent line of the function $f\left(x\right)$ is given as $\left(3-4x\right)$, the function $f^{\prime }\left(x\right)$ is $\left(3-4x\right)$.

$f^{\prime }\left(x\right)=\left(3-4x\right)$

Calculate the function $f\left(x\right)$ from the tangent function $f^{\prime }\left(x\right)$ as follows.

Calculation of $f\left(x\right)$:

Rewrite the function $f^{\prime }\left(x\right)=\left(3-4x\right)$ as follows.

$f^{\prime }\left(x\right)=-4x^1+3x^0$ (1)

From the antiderivative function formula, the antiderivative for the function in equation (1) is written as follows.

$\begin{array}{c}f\left(x\right)=-4\left(\frac{x^{1+1}}{1+1}\right)+3\left(\frac{x^{0+1}}{0+1}\right)+C\\ =-4\left(\frac{x^2}{2}\right)+3x+C\end{array}$

$f\left(x\right)=-2x^2+3x+C$ (2)

As the function $f\left(x\right)$ passes through the point $\left(2,5\right)$, the value of $f\left(2\right)$ is 5.

As $f\left(2\right)=5$, substitute 2 for x in equation (2),

$\begin{array}{c}f\left(2\right)=-2{\left(2\right)}^2+3\left(2\right)+C\\ =-8+6+C\\ =-2+C\end{array}$

Rewrite the expression as follows.

$C=f\left(2\right)+2$

Substitute 5 for $f\left(2\right)$,

$\begin{array}{c}C=5+2\\ =7\end{array}$

Substitute 7 for C in equation (2),

$f\left(x\right)=-2x^2+3x+7$ (3)

Calculation of the value $f\left(1\right)$:

Substitute 1 for x in equation (3),

$\begin{array}{c}f\left(1\right)=-2{\left(1\right)}^2+3\left(1\right)+7\\ =-2+3+7\\ =8\end{array}$

Thus, the value of $f\left(1\right)$ is $\underset{\_}{8}$.