If fx-ax2-bx-c then show that limh - 0fx-h-fxh-2ax-b-

Given, f(x)=ax^2+bx+c .Now f(x+h) f(x) =a{(x+h)^2 x^2}+b{(x+h) x}+c c =2ahx+ah^2+bh .Now, lim_h → 0f(x+h) f(x)h =lim_h → 02ahx+ah^2+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Limit

If fx=ax2+b...

Question

If $f (x) = a x^{2} + b x + c$ then show that $lim h \to 0 \frac{f (x + h) - f (x)}{h} = 2 a x + b$ .

Open in App

Solution

Given,
$f (x) = a x^{2} + b x + c$ .
Now $f (x + h) - f (x)$
$= a {(x + h)^{2} - x^{2}} + b {(x + h) - x} + c - c$
$= 2 a h x + a h^{2} + b h$ .
Now,
$lim h \to 0 \frac{f (x + h) - f (x)}{h}$
$= lim h \to 0 \frac{2 a h x + a h^{2} + b h}{h}$
$= lim h \to 0 (2 a x + a h + b)$
$= 2 a x + b$ .

Suggest Corrections

Similar questions

Left hand derivative and right hand derivative of a function

f (x)

at a point

x = a

are defined as

f^{'} (a^{-}) = lim h \to 0^{+} \frac{f (a) - f (a - h)}{h} = lim h \to 0^{-} \frac{f (a) - f (a - h)}{h} = lim x \to a^{+} \frac{f (a) - f (x)}{a - x}

respectively

Let

f

be a twice differentiable function. We also know that derivative of an even function is odd function and derivative of an odd function is even function.

The statement

lim h \to 0 \frac{f (- x) - f (- x - h)}{h} = lim h \to 0 \frac{f (x) - f (x - h)}{- h}

implies that for all x

ϵ R

Q. Compute the value of

θ

in the first mean value theorem

f (x + h) = f (x) + h f^{'} (x + θ h)

f (x) = a x^{2} + b x + c .

$f (x) = \frac{a x^{2} + b}{x^{2} + 1}, {lim}_{x \to 0} f (x) = 1 a n d {lim}_{x \to \infty} f (x) = 1, t h e n p r o v e t h a t f (- 2) = f (2) = 1.$

Q. If the given figure shows the graph of

f (x) = a x^{2} + b x + c,

then

Q. Let

f (x) = lim h \to 0 \frac{(sin (x + h))^{ln (x + h)} - (sin x)^{ln x}}{h}

then

f (\frac{π}{2})

Join BYJU'S Learning Program

If f(x)=ax2+bx+c then show that limh→0f(x+h)−f(x)h=2ax+b.

Given,f(x)=ax2+bx+c.Now f(x+h)−f(x)=a{(x+h)2−x2}+b{(x+h)−x}+c−c=2ahx+ah2+bh.Now,limh→0f(x+h)−f(x)h=limh→02ahx+ah2+bhh=limh→0(2ax+ah+b)=2ax+b.

If $f (x) = a x^{2} + b x + c$ then show that $lim h \to 0 \frac{f (x + h) - f (x)}{h} = 2 a x + b$ .

Given,
$f (x) = a x^{2} + b x + c$ .
Now $f (x + h) - f (x)$
$= a {(x + h)^{2} - x^{2}} + b {(x + h) - x} + c - c$
$= 2 a h x + a h^{2} + b h$ .
Now,
$lim h \to 0 \frac{f (x + h) - f (x)}{h}$
$= lim h \to 0 \frac{2 a h x + a h^{2} + b h}{h}$
$= lim h \to 0 (2 a x + a h + b)$
$= 2 a x + b$ .