in case you ever took precalculus or another math type which covers properties of applications, you study a thank you to ensure the area of the function. between the pink flags to look out for is once you divide by making use of 0. on your expression, while x = 3, the denominator is 0. for this reason, the area is each type beside 3. regrettably, the variety you attempt to plug in (x = 3) is the only type that doesn't artwork in this function. to work out this usual hand, you0 can graph this function on a TI-80 3. in case you zoom in close to the graph at x = 3, you will see that there is a sparkling spot there! it somewhat is via the fact, as reported above, there basically isn't a value of the expression at x = 3. you may say, nicely curiously like the respond would desire to be 6, finding on the graph. this thought-approximately what the respond "would desire to be" is what limits are all approximately. The values of the function on the left and suited of x = 3 all pass in direction of 6 as you get closer and closer. So we are saying the cut back as x is going to 3 is 6. So even with the certainty that it's not technically the respond, 6 is your terrific selection. 0 isn't ideal in any experience. the very terrific answer is to assert that the expression is undefined at x = 3. This problem illustrates why 0/0 is termed indeterminate. in this problem, 0/0 in a manner equals 6. the thought 0/0 can equivalent something is honestly the essence of calculus.
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Verified answer
ƒ(x) = 8x³ - 7x² + 2√x - 53
=> differentiate with respect to x
=> f'(x) = d(8x³)/dx - d(7x²)/dx + d(2√x)/dx - d(53)/dx
=> f'(x) = 8d(x³)/dx - 7d(x²)/dx + 2d(√x)/dx - d(53)/dx
=> f'(x) = 8(3x²) - 7(2x) + 2(1/2√x) - 0
=> f'(x) = 24x² - 14x + 1/√x
in case you ever took precalculus or another math type which covers properties of applications, you study a thank you to ensure the area of the function. between the pink flags to look out for is once you divide by making use of 0. on your expression, while x = 3, the denominator is 0. for this reason, the area is each type beside 3. regrettably, the variety you attempt to plug in (x = 3) is the only type that doesn't artwork in this function. to work out this usual hand, you0 can graph this function on a TI-80 3. in case you zoom in close to the graph at x = 3, you will see that there is a sparkling spot there! it somewhat is via the fact, as reported above, there basically isn't a value of the expression at x = 3. you may say, nicely curiously like the respond would desire to be 6, finding on the graph. this thought-approximately what the respond "would desire to be" is what limits are all approximately. The values of the function on the left and suited of x = 3 all pass in direction of 6 as you get closer and closer. So we are saying the cut back as x is going to 3 is 6. So even with the certainty that it's not technically the respond, 6 is your terrific selection. 0 isn't ideal in any experience. the very terrific answer is to assert that the expression is undefined at x = 3. This problem illustrates why 0/0 is termed indeterminate. in this problem, 0/0 in a manner equals 6. the thought 0/0 can equivalent something is honestly the essence of calculus.
Derivative will take this form.
A=Cx^n
A'=Cnx^(n-1)
√x=x^(1/2)
derivative = (1/2)x^(-1/2) = 1/(2√x)
So we have f'(x)=24x^2-14x+1/√x
we are using power rule here that is:
f(x)=x^n=nx^n-1
differentiating this problem would yield to:
24x^2-14x+x^-1.5
d/dx (f(x)) = 24 x² - 14 x + (1/√x)
Simplify to the extent you want.
d/dx (x^n) = n [x^(n-1)]
and
d/dx (constant) = 0
Only Albert Einstein could work that out
f ` (x) = 24x² - 14 x + 1/√x