Question

Find the maximum or minimum value of the quadratic expression $$2x-7-5x^2$$.

Answer 1

Hint: We know that the minimum or maximum value of a quadratic expression $$y=a{x}^2+bx+c$$ is $$\left({\displaystyle \frac{4ac-b^2}{4a}}\right)$$ at $$x={\displaystyle \frac{-b}{2a}}$$. If $a<0$, then the quadratic expression will have a maximum value. We will compare $$2x-7-5x^2$$ with $$a{x}^2+bx+c$$. Now, we will get the values of a, b and c. With these values of a, b and c, we will find the minimum or maximum values of $$2x-7-5x^2$$.

Complete step-by-step solution -
Before solving the question, we should whether a quadratic expression will have maximum value (or) minimum value.
For a quadratic expression $$y=a{x}^2+bx+c$$, if $a<0$ then the quadratic expression will have maximum value. The maximum value of $$y=a{x}^2+bx+c$$ obtains at $$x={\displaystyle \frac{-b}{2a}}$$. The maximum value of quadratic expression is $$\left({\displaystyle \frac{4ac-b^2}{4a}}\right)$$.
In the similar way, if $a>0$ then the quadratic expression will have minimum value. The minimum value of $$y=a{x}^2+bx+c$$ obtains at $$x={\displaystyle \frac{-b}{2a}}$$. The minimum value of quadratic expression is $$\left({\displaystyle \frac{4ac-b^2}{4a}}\right)$$.
The given expression in this question is $$2x-7-5x^2$$.
Let us assume $$y=2x-7-5x^2$$
By rewriting the quadratic expression,
$$y=-5x^2+2x-7$$
Now we should compare $$y=-5x^2+2x-7$$ with $$y=a{x}^2+bx+c$$.
$$\begin{array}{rl}& a=-5.....(1)\\ & b=2........(2)\\ & c=-7......(3)\end{array}$$
We know that if $a<0$, then the quadratic expression will have a maximum value.
From equation (1), it is clear that the value of a for $$y=-5x^2+2x-7$$ is less than zero.
So, the quadratic expression $$-5x^2+2x-7$$ will have a maximum value.

We know that the minimum value for a quadratic expression will obtain at $$x={\displaystyle \frac{-b}{2a}}$$.
From equation (2) and equation (3), the maximum value of quadratic expression will obtain at
$$x={\displaystyle \frac{-2}{2(-5)}}={\displaystyle \frac{-2}{-10}}={\displaystyle \frac{2}{10}}={\displaystyle \frac{1}{5}}.....(5)$$.
We know that the maximum value of quadratic expression is $$\left({\displaystyle \frac{4ac-b^2}{4a}}\right)$$.
So, the maximum value of quadratic expression is
$${\displaystyle \frac{4ac-b^2}{4a}}={\displaystyle \frac{4(-5)(-7)-{(2)}^2}{4(-5)}}={\displaystyle \frac{4(35)-4}{-20}}={\displaystyle \frac{140-4}{-20}}={\displaystyle \frac{136}{-20}}={\displaystyle \frac{-68}{10}}={\displaystyle \frac{-34}{5}}$$.
Hence, the maximum value of $$2x-7-5x^2$$ is $${\displaystyle \frac{-34}{5}}$$.

Note: There is an alternative method to solve this problem.
A function f(x) is said to have a maximum or minimum value at the value of x where $f^{\prime }(x)=0$.
The value of x where f`(x)=0 is said to have a maximum value if $f”(x)<0$.
The value of x where f`(x)=0 is said to have a minimum value if $f”(x)>0$.
Let us assume $$f(x)=2x-7-5x^2......(1)$$ .
$$\Rightarrow f^{\prime }(x)={\displaystyle \frac{d}{dx}}(2x-7-5x^2)=2-10x$$
We have to find the value of x where f`(x) is equal to 0.
$$\begin{array}{rl}& f^{\prime }(x)=2-10x=0\\ & \Rightarrow 10x=2\\ & \Rightarrow x={\displaystyle \frac{1}{5}}......(2)\end{array}$$
At $$x={\displaystyle \frac{1}{5}}$$, $$f(x)=2x-7-5x^2$$ will have a maximum (or) minimum value.
$$f”(x)={\displaystyle \frac{d^2}{d{x}^2}}(2x-7-5x^2)={\displaystyle \frac{d}{dx}}{\displaystyle \frac{d}{dx}}(2x-7-5x^2)={\displaystyle \frac{d}{dx}}(2-10x)=-10$$
As $f^{″}(x)<0$, so f(x) will have maximum value at $$x={\displaystyle \frac{1}{5}}$$.
So, substitute equation (2) in equation (1).
$$f(x)=2\left({\displaystyle \frac{1}{5}}\right)-7-5{\left({\displaystyle \frac{1}{5}}\right)}^2={\displaystyle \frac{-34}{5}}$$.
Hence, the maximum value of $$2x-7-5x^2$$ is equal to $${\displaystyle \frac{-34}{5}}$$.