Given:
The function is
[tex]f(x)=-4x^2+13x-7[/tex]
To find:
The interval on which the function is increasing.
Solution:
We have,
[tex]f(x)=-4x^2+13x-7[/tex]
Differentiate with respect to x.
[tex]f'(x)=-4(2x)+13(1)-(0)[/tex]
[tex]f'(x)=-8x+13[/tex]
Equate f'(x)=0.
[tex]-8x+13=0[/tex]
[tex]-8x=-13[/tex]
[tex]x=\dfrac{-13}{-8}[/tex]
[tex]x=1.625[/tex]
The point x=1.625 divides the number line in two parts [tex](-\infty, 1.625)\text{ and }(1.625, \infty)[/tex].
f'(x) is positive for [tex](-\infty, 1.625)[/tex]. It means the function is increasing on this interval.
f'(x) is negative for [tex](1.625,\infty)[/tex]. It means the function is decreasing on this interval.
At x=1.625, f'(x) is 0. It is turning point so it will included in both intervals.
Therefore, the function is increasing on [tex](-\infty, 1.625][/tex].
